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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
int dir[8][2] = {-1, 0, 1, 0, 0, -1, 0, 1, -1, -1, -1, 1, 1 - 1, 1, 1};
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = 1LL * a * b % MOD;
}
template <typename T>
inline bool IsPri(T x) {
  if (x < 2) return false;
  for (T i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
template <typename T>
inline T Gcd(T a, T b) {
  while (b) {
    T t = b;
    b = a % b;
    a = t;
  }
  return a;
}
template <typename T>
inline int BitCnt(T x) {
  int cnt = 0;
  while (x) ++cnt, x &= x - 1;
  return cnt;
}
template <typename T>
inline T MDis(T x1, T y1, T x2, T y2) {
  return abs(x1 - x2) + abs(y1 - y2);
}
template <typename T>
inline T MDis(T x1, T y1, T z1, T x2, T y2, T z2) {
  return abs(x1 - x2) + abs(y1 - y2) + abs(z1 - z2);
}
inline double EDis(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
inline double EDis(double x1, double y1, double z1, double x2, double y2,
                   double z2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) +
              (z1 - z2) * (z1 - z2));
}
inline long long Pow(long long a, long long n) {
  long long t = 1;
  a %= MOD;
  while (n > 0) {
    if (n & 1) t = t * a % MOD;
    a = a * a % MOD, n >>= 1;
  }
  return t % MOD;
}
inline int read() {
  static char buf[1000000], *p1 = buf, *p2 = buf;
  register int x = false;
  register char ch =
      p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
          ? EOF
          : *p1++;
  ;
  register bool sgn = false;
  while (ch != '-' && (ch < '0' || ch > '9'))
    ch = p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
  ;
  if (ch == '-')
    sgn = true,
    ch = p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
  ;
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48),
    ch = p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
  ;
  return sgn ? -x : x;
}
const int N = 4e5 + 100;
int n, m, k, Q;
vector<int> e[N];
int rt[N], vis[N];
int find(int x) {
  if (rt[x] != x) rt[x] = find(rt[x]);
  return rt[x];
}
void merge(int x, int y) { rt[find(x)] = find(y); }
queue<pair<int, int> > q;
void BFS() {
  while (!q.empty()) {
    int u = q.front().first, d = q.front().second;
    q.pop();
    for (int v : e[u])
      if (vis[v])
        merge(u, v);
      else if (d < k)
        q.push({v, d + 1}), vis[v] = 1;
  }
}
int dep[N], fz[N][20];
void DFS(int x, int f) {
  dep[x] = dep[f] + 1, fz[x][0] = f;
  for (int i = 1; i < 20; ++i) fz[x][i] = fz[fz[x][i - 1]][i - 1];
  for (int y : e[x])
    if (y != f) DFS(y, x);
}
int Up(int x, int k) {
  for (int i = 0; i < 20; ++i)
    if (k >> i & 1) x = fz[x][i];
  return x;
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  x = Up(x, dep[x] - dep[y]);
  if (x == y) return x;
  for (int i = 19; i >= 0; --i)
    if (fz[x][i] != fz[y][i]) x = fz[x][i], y = fz[y][i];
  return fz[x][0];
}
int Move(int x, int y, int k) {
  int a = LCA(x, y);
  if (dep[x] - dep[a] >= k) return Up(x, k);
  return Up(y, dep[x] + dep[y] - dep[a] * 2 - k);
}
int main() {
  for (int i = 0; i < N; ++i) rt[i] = i;
  cin >> n >> k >> m;
  int u, v;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    e[u].push_back(n + i), e[v].push_back(n + i);
    e[n + i].push_back(u), e[n + i].push_back(v);
  }
  for (int i = 1; i <= m; ++i) scanf("%d", &u), q.push({u, 0}), vis[u] = u;
  BFS(), DFS(1, 0);
  cin >> Q;
  while (Q--) {
    scanf("%d%d", &u, &v);
    int a = LCA(u, v);
    if (dep[u] + dep[v] - dep[a] * 2 <= k * 2)
      puts("YES");
    else {
      int uu = Move(u, v, k), vv = Move(v, u, k);
      puts(find(uu) == find(vv) ? "YES" : "NO");
    }
  }
  return 0;
}