// https://yukicoder.me/problems/no/2319
// really cool problem
//
// Solution 0. use bitsets and bruteforce
//
// 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. sqrt over queries
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;

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

// signed main2() {
//   cout << "sol 1\n";
//   freopen("/Users/drymerge/CLionProjects/DEBUG/a.in", "r", stdin);
//   freopen("/Users/drymerge/a.out", "w", stdout);
//   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;
//   }
// }

signed main() {
  // cout << "sol2\n";
  // freopen("/Users/drymerge/CLionProjects/DEBUG/a.in", "r", stdin);
  // freopen("/Users/drymerge/b.out", "w", stdout);
  int N, M, Q;
  cin >> N >> M;
  
  vector<int> world(N+1); 
  for (int i = 1; i <= N; ++i) 
    cin >> world[i];

  vector<unordered_set<int>> friends(N+1);
  vector<array<int,2>> friendships(M);
  unordered_map<int, unordered_map<int, int>> initial_count;
  // initial_count[i][j] = # of friends of i in world j
  for (auto& [a,b] : friendships) {
    cin >> a >> b;
    friends[a].insert(b);
    friends[b].insert(a);
  }

  auto reset_map = [&] {
    initial_count.clear();
    for (auto [a,b] : friendships) {
      ++initial_count[a][world[b]];
      ++initial_count[b][world[a]];
    }
  };

  auto is_friends = [&] (int i, int j) -> bool {
    if (friends[i].size() > friends[j].size()) swap(i, j);
    return friends[i].count(j);
  };

  cin >> Q;
  int K = sqrt(Q);
  int q = 0;
  vector<array<int,4>> succeeded;
  while (q < Q) {
    reset_map();
    for (int k = 0; k < K && q < Q; ++k, ++q) {
      int x, y;
      cin >> x >> y;
      if (world[x] == world[y]) {
        cout << "No\n";
        continue;
      }

      int start = initial_count[x][world[y]];
      for (auto [xl,yl,wx,wy] : succeeded) {
        if (wx == world[y] && is_friends(x, xl))
          --start;
        if (wy == world[y] && is_friends(x, xl))
          ++start;
      }
      if (start > 0) {
        cout << "Yes\n";
        succeeded.push_back( {x,y,world[x],world[y]} );
        world[x] = world[y];
      }
      else cout << "No\n";
    }
    succeeded.clear();
  }
  return 0;
}