Problem: Bessie is planning a vacation! In Cow-lifornia, there are n cities, with n-1 bidirectional roads connecting them. It is guaranteed that one can reach any city from any other city. 

Bessie is considering v possible vacation plans, with the i-th one consisting of a start city a_i and destination city b_i.

It is known that only r of the cities have rest stops. Bessie gets tired easily, and cannot travel across more than k consecutive roads without resting. In fact, she is so desperate to rest that she may travel through the same city multiple times in order to do so.

For each of the vacation plans, does there exist a way for Bessie to travel from the starting city to the destination city?

Input

The first line contains three integers n, k, and r (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k,r ≤ n) — the number of cities, the maximum number of roads Bessie is willing to travel through in a row without resting, and the number of rest stops.

Each of the following n-1 lines contain two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), meaning city x_i and city y_i are connected by a road. 

The next line contains r integers separated by spaces — the cities with rest stops. Each city will appear at most once.

The next line contains v (1 ≤ v ≤ 2 ⋅ 10^5) — the number of vacation plans.

Each of the following v lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the start and end city of the vacation plan. 

Output

If Bessie can reach her destination without traveling across more than k roads without resting for the i-th vacation plan, print YES. Otherwise, print NO.

Examples

Input


6 2 1
1 2
2 3
2 4
4 5
5 6
2
3
1 3
3 5
3 6


Output


YES
YES
NO


Input


8 3 3
1 2
2 3
3 4
4 5
4 6
6 7
7 8
2 5 8
2
7 1
8 1


Output


YES
NO

Note

The graph for the first example is shown below. The rest stop is denoted by red.

For the first query, Bessie can visit these cities in order: 1, 2, 3.

For the second query, Bessie can visit these cities in order: 3, 2, 4, 5. 

For the third query, Bessie cannot travel to her destination. For example, if she attempts to travel this way: 3, 2, 4, 5, 6, she travels on more than 2 roads without resting.

<image>

The graph for the second example is shown below. 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
long long n, k, r;
const long long N = 400100;
vector<long long> g[N];
long long L[N];
const long long MAXL = 20;
long long ans[N][MAXL];
long long pai[N];
long long d[N];
long long f(long long x) { return (x == pai[x] ? x : pai[x] = f(pai[x])); }
void join(long long a, long long b) {
  if (f(a) == f(b)) return;
  pai[f(a)] = f(b);
}
void dfs(long long v, long long p = -1) {
  for (long long to : g[v]) {
    if (to == p) continue;
    ans[to][0] = v;
    L[to] = L[v] + 1;
    dfs(to, v);
  }
}
long long LCA(long long a, long long b) {
  if (L[a] < L[b]) swap(a, b);
  for (long long j = MAXL - 1; j >= 0; j--) {
    if (L[a] - (1 << j) >= L[b]) {
      a = ans[a][j];
    }
  }
  if (a == b) return a;
  for (long long j = MAXL - 1; j >= 0; j--) {
    if (ans[a][j] != ans[b][j]) {
      a = ans[a][j], b = ans[b][j];
    }
  }
  return ans[a][0];
}
long long lca;
long long go(long long ini, long long fim, long long k) {
  if (L[ini] > L[lca]) {
    for (long long j = MAXL - 1; j >= 0; j--) {
      if (L[ini] - (1 << j) >= L[lca] and k >= (1 << j)) {
        k -= (1 << j);
        ini = ans[ini][j];
      }
    }
  }
  if (ini != lca) return ini;
  k = L[fim] - L[lca] - k;
  for (long long j = MAXL - 1; j >= 0; j--) {
    if (k >= (1 << j)) {
      k -= (1 << j);
      fim = ans[fim][j];
    }
  }
  return fim;
}
long long ok(long long a, long long b, long long k) {
  lca = LCA(a, b);
  if (L[a] + L[b] - 2 * L[lca] <= 2 * k) return true;
  if (f(go(a, b, k)) == f(go(b, a, k))) return true;
  return false;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k >> r;
  long long cur = n + 1;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].push_back(cur);
    g[cur].push_back(a);
    g[cur].push_back(b);
    g[b].push_back(cur);
    cur++;
  }
  queue<long long> q;
  n = cur - 1;
  for (long long i = 1; i <= n; i++) {
    pai[i] = i, d[i] = -1;
  }
  for (long long i = 1; i <= r; i++) {
    long long x;
    cin >> x;
    d[x] = 0;
    q.push(x);
  }
  while (!q.empty()) {
    long long cur = q.front();
    q.pop();
    if (d[cur] >= k) continue;
    for (long long to : g[cur]) {
      join(to, cur);
      if (d[to] == -1) {
        d[to] = d[cur] + 1;
        q.push(to);
      }
    }
  }
  ans[1][0] = 1;
  dfs(1);
  for (long long j = 1; j < MAXL; j++) {
    for (long long i = 1; i <= n; i++) {
      ans[i][j] = ans[ans[i][j - 1]][j - 1];
    }
  }
  long long qry;
  cin >> qry;
  while (qry--) {
    long long a, b;
    cin >> a >> b;
    if (ok(a, b, k))
      cout << "YES"
           << "\n";
    else
      cout << "NO"
           << "\n";
  }
}