// https://yukicoder.me/problems/no/2319
// really cool problem
//
// Solution 0. use bitsets and bruteforce (cheating)
//
// Solution 1. sqrt over nodes
// let K = sqrt(M)
// a person is "few" if it has < K friends and "many" if >= K
// there are at most 2M / K ~= K people that are "many"
// to solve a query where x is few, brute force it
// to solve a query where x is many, consider its friends that are:
//     a. many - you can check ALL many's for the condition
//     b. few  - store this as map f[i][j] = # of few friends of the many i in world j
// update naively, it works in O(Q*sqrt M + N*sqrt M + M)
//
// Solution 2. ft. Jerry -> sqrt over queries
// for each chunk, recompute a map of for each person i, how many friends they have in world j
// then for each query, loop over the previous queries in the chunk and simulate. if by the end,
// you still have at least one friend in world[y], ur good
// the map computations take M sqrt(Q) time total, the rest is Q sqrt(Q)
#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;

using HashSet = unordered_set<int>;
using FewCounter = unordered_map<int, vector<int>>;

// signed main2() {
//   cin.tie(0)->sync_with_stdio(0);
//   int N, M, Q;
//   cin >> N >> M;
//   int K = sqrt(M);
//   vector<int>  world(N+1);  // world[i] = the world that person i is in
//   FewCounter   num_few_of;  // num_few_of[i].at(w) = # friends of i in world j with < K friends
//   vector<int>  many;        // many = [those with >= K friends]
//   vector<HashSet> friends(N+1);

//   auto is_many = [&] (int i) {
//     return friends[i].size() >= K;
//   };

//   for (int i = 1; i <= N; ++i)
//     cin >> world[i];

//   for (int i = 1; i <= M; ++i) {
//     int a, b;
//     cin >> a >> b;
//     friends[a].insert(b);
//     friends[b].insert(a);
//   }

//   for (int i = 1; i <= N; ++i) {
//     if (!is_many(i)) continue;
//     many.push_back(i);
//     num_few_of[i].assign(N+1, 0);
//     for (int f : friends[i])
//       if (!is_many(f)) {
//         int w = world[f];
//         ++num_few_of[i].at(w);
//       }
//   }

//   auto solve_many = [&] (int x, int y) {
//     int new_world = world[y];
//     if (num_few_of[x].at(new_world) > 0)
//       goto success;

//     for (int m : many)
//       if (friends[x].count(m) && world[m] == new_world)
//         goto success;

//     return "No\n";

//     success:
//     world[x] = new_world;
//     return "Yes\n";
//   };

//   auto solve_few = [&] (int x, int y) {
//     for (int f : friends[x])
//       if (world[f] == world[y])
//         goto success;

//     return "No\n";

//     success:
//     int old_world = world[x];
//     int new_world = world[y];
//     for (int f : friends[x]) {
//       if (!is_many(f)) continue;
//       --num_few_of[f].at(old_world);
//       ++num_few_of[f].at(new_world);
//     }
//     world[x] = new_world;
//     return "Yes\n";
//   };

//   cin >> Q;
//   while (Q--) {
//     int x, y;
//     cin >> x >> y;
//     if (world[x] == world[y]) {
//       cout << "No\n";
//       continue;
//     }
//     auto ans = is_many(x) ? solve_many(x,y) : solve_few(x,y);
//     cout << ans;
//   }
// }

// SOLUTION 2: TLEs sadly. could not get it to pass
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
struct chash {
    const int RANDOM = (long long)(make_unique<char>().get()) ^ chrono::high_resolution_clock::now().time_since_epoch().count();
    static unsigned long long hash_f(unsigned long long x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    static unsigned hash_combine(unsigned a, unsigned b) { return a * 31 + b; }
    int operator()(int x) const { return hash_f(x)^RANDOM; }
};
template <typename K, typename V>
using HashTable = gp_hash_table<K,V,chash>;

// template <typename K, typename V>
// using HashTable = unordered_map<K,V>;
constexpr size_t MXN = 20001;
constexpr size_t MXM = 400000;
// constexpr size_t MXN = 100;
// constexpr size_t MXM = 100;
constexpr size_t SQR = 450;

#define all(x) begin(x), end(x)

signed main() {
    cin.tie(0)->sync_with_stdio(0);
    int N, M, Q;
    cin >> N >> M;

    short world[MXN];
    bitset<MXN> friends[MXN];
    vector<short> friendss[MXN];
    short friendships[MXM];
    HashTable<int, short> initial_count[MXN];
    // short initial_count[MXN][SQR];
    short X[SQR], Y[SQR];

    vector<int> whos_involved;

    // gp_hash_table<int,short> initial_count[MXN];
    // initial_count[i][j] = # of friends of i in world j

    for (int i = 1; i <= N; ++i)
        cin >> world[i];

    for (int i = 0; i < M; ++i) {
        short a, b;
        cin >> a >> b;
        friendss[a].push_back(b);
        friendss[b].push_back(a);
        friendships[2*i] = a, friendships[2*i+1] = b;
        friends[a][b] = true;
        friends[b][a] = true;
    }

    auto reset_map = [&] {
        sort( all(whos_involved) );
        whos_involved.resize(unique(all(whos_involved)) - begin(whos_involved));

        for (int c : whos_involved) {
            if (!initial_count[c].empty())
                initial_count[c].clear();
            for (int f : friendss[c]) {
                ++initial_count[c][world[f]];
            }
        }

        // for (int i = 0; i < M; ++i) {
        //   short a = friendships[2*i];
        //   short b = friendships[2*i+1];
        //   ++initial_count[a][world[b]];
        //   ++initial_count[b][world[a]];
        // }
    };

    cin >> Q;
    int K = sqrt(Q);
    int q = 0;
    HashTable<int,vector<int>> succeeded_wx, succeeded_wy;

    while (q < Q) {
        int last = min(Q, q + K);
        for (int k = q; k < last; ++k) {
            cin >> X[k-q] >> Y[k-q];
            whos_involved.push_back(X[k-q]);
            whos_involved.push_back(Y[k-q]);
        }
        reset_map();
        for (int k = 0; k < K && q < Q; ++k, ++q) {
            short x = X[k], y = Y[k];
            if (world[x] == world[y]) {
                cout << "No\n";
                continue;
            }

            int start = initial_count[x][world[y]];
            for (int xl : succeeded_wx[ world[y] ]) {
                start -= (friends[x][xl]);
            }
            for (int xl : succeeded_wy[ world[y] ]) {
                start += (friends[x][xl]);
            }
            if (start > 0) {
                cout << "Yes\n";
                succeeded_wx[world[x]].push_back(x);
                succeeded_wy[world[y]].push_back(x);
                world[x] = world[y];
            }
            else cout << "No\n";
        }
        succeeded_wx.clear();
        succeeded_wy.clear();
        whos_involved.clear();
    }
    return 0;
}