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;
constexpr int maxn = 5e5 + 100, maxdep = 20;
struct IO {
 private:
  static constexpr int ibufsiz = 1 << 20;
  char ibuf[ibufsiz + 1], *inow = ibuf, *ied = ibuf;
  static constexpr int obufsiz = 1 << 20;
  char obuf[obufsiz + 1], *onow = obuf;
  const char *oed = obuf + obufsiz;

 public:
  char getchar() {
    if (inow == ied) {
      ied = ibuf + sizeof(char) * fread(ibuf, sizeof(char), ibufsiz, stdin);
      *ied = '\0';
      inow = ibuf;
    }
    return *inow++;
  }
  template <typename T>
  void read(T &x) {
    static bool flg;
    flg = 0;
    x = 0;
    char c = getchar();
    while (!isdigit(c)) flg = c == '-' ? 1 : flg, c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    if (flg) x = -x;
  }
  template <typename T, typename... Y>
  void read(T &x, Y &...X) {
    read(x);
    read(X...);
  }
  int readi() {
    static int res;
    read(res);
    return res;
  }
  long long readll() {
    static long long res;
    read(res);
    return res;
  }
  void flush() {
    fwrite(obuf, sizeof(char), onow - obuf, stdout);
    fflush(stdout);
    onow = obuf;
  }
  void putchar(const char c) {
    *onow++ = c;
    if (onow == oed) {
      fwrite(obuf, sizeof(char), obufsiz, stdout);
      onow = obuf;
    }
  }
  template <typename T>
  void write(T x, const char split = '\0') {
    static unsigned char buf[32];
    if (x < 0) putchar('-'), x = -x;
    int p = 0;
    do {
      buf[++p] = x % 10;
      x /= 10;
    } while (x);
    for (int i = p; i >= 1; i--) putchar(buf[i] + '0');
    if (split != '\0') putchar(split);
  }
  void puts(char const *str, const char split = '\n') {
    while (*str) putchar(*(str++));
    putchar(split);
  }
  void lf() { putchar('\n'); }
  ~IO() { fwrite(obuf, sizeof(char), onow - obuf, stdout); }
} io;
vector<int> G[maxn];
void addedge(const int u, const int v) { G[u].push_back(v); }
int n, lim, r, tot, mark[maxn], f[maxn], dep[maxn], faz[maxn][maxdep + 1],
    bel[maxn];
void dfs(const int u) {
  for (int i = 1; i <= maxdep; i++) faz[u][i] = faz[faz[u][i - 1]][i - 1];
  dep[1] = 1;
  for (int v : G[u]) {
    if (v == faz[u][0]) continue;
    faz[v][0] = u;
    dep[v] = dep[u] + 1;
    dfs(v);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = maxdep; i >= 0; i--)
    if (dep[faz[x][i]] >= dep[y]) x = faz[x][i];
  if (x == y) return x;
  for (int i = maxdep; i >= 0; i--)
    if (faz[x][i] != faz[y][i]) x = faz[x][i], y = faz[y][i];
  return faz[x][0];
}
int find(const int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void merge(const int a, const int b) {
  const int x = find(a), y = find(b);
  f[x] = y;
}
void init() {
  queue<int> q;
  for (int i = 1; i <= tot; i++)
    if (mark[i]) q.push(i), bel[i] = i;
  for (int i = 1; i <= tot; i++) f[i] = i;
  for (int i = 1; i <= lim; i++) {
    int mem = q.size();
    while (mem--) {
      const int u = q.front();
      q.pop();
      for (int v : G[u]) {
        if (!bel[v])
          bel[v] = bel[u], f[v] = find(u), q.push(v);
        else if (bel[v] != bel[u])
          merge(u, v);
      }
    }
  }
}
int walk(const int x, const int y) {
  auto kth = [](int u, const int k) {
    for (int i = 0; i <= maxdep; i++)
      if ((k >> i) & 1) u = faz[u][i];
    return u;
  };
  const int l = lca(x, y);
  if (lim <= dep[x] - dep[l]) return kth(x, lim);
  return kth(y, (dep[y] - dep[l]) - (lim - (dep[x] - dep[l])));
}
int main() {
  io.read(n, lim, r);
  tot = n;
  for (int u, v, i = 1; i < n; i++) {
    io.read(u, v);
    tot++;
    addedge(u, tot);
    addedge(tot, u);
    addedge(v, tot);
    addedge(tot, v);
  }
  dfs(1);
  for (int i = 1; i <= r; i++) mark[io.readi()] = 1;
  init();
  const int q = io.readi();
  for (int x, y, i = 1; i <= q; i++) {
    io.read(x, y);
    const int l = lca(x, y);
    if (dep[x] + dep[y] - 2 * dep[l] <= 2 * lim) {
      puts("YES");
      continue;
    }
    if (find(walk(x, y)) == find(walk(y, x))) {
      puts("YES");
      continue;
    }
    puts("NO");
  }
  return 0;
}