// 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;

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename K, typename V>
using HashTable = gp_hash_table<K,V>;

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

signed main() {
  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: took forever to make it pass :/

// constexpr size_t MXN = 20001;
// constexpr size_t MXM = 400000;
// 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];
//     // initial_count[i][j] = # of friends of i in world j
//     short X[SQR], Y[SQR];

//     vector<int> whos_involved;

//     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]];
//             }
//         }
//     };

//     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;
// }