Problem: You have a rooted tree consisting of n vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to n. Then we represent the color of vertex v as cv. The tree root is a vertex with number 1.

In this problem you need to answer to m queries. Each query is described by two integers vj, kj. The answer to query vj, kj is the number of such colors of vertices x, that the subtree of vertex vj contains at least kj vertices of color x.

You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).

Input

The first line contains two integers n and m (2 ≤ n ≤ 105; 1 ≤ m ≤ 105). The next line contains a sequence of integers c1, c2, ..., cn (1 ≤ ci ≤ 105). The next n - 1 lines contain the edges of the tree. The i-th line contains the numbers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the vertices connected by an edge of the tree.

Next m lines contain the queries. The j-th line contains two integers vj, kj (1 ≤ vj ≤ n; 1 ≤ kj ≤ 105).

Output

Print m integers — the answers to the queries in the order the queries appear in the input.

Examples

Input

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


Output

2
2
1
0
1


Input

4 1
1 2 3 4
1 2
2 3
3 4
1 1


Output

4

Note

A subtree of vertex v in a rooted tree with root r is a set of vertices {u : dist(r, v) + dist(v, u) = dist(r, u)}. Where dist(x, y) is the length (in edges) of the shortest path between vertices x and y.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void remax(T &A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T &A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  for (int i = s.size() - 1; i >= 0; --i) r += (s[i] - '0') * p, p *= 10;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, long long power) {
  if (power < 0) return 0;
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
long long PowerMod(long long base, long long power, long long mod) {
  if (!power) return 1;
  if (power & 1) return (base * PowerMod(base, power - 1, mod)) % mod;
  return PowerMod((base * base) % mod, power >> 1, mod);
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
int Count(long long mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
struct query {
  int l;
  int r;
  int k;
  int ind;
  const bool operator<(const query &rhs) const;
} q[100000];
vector<int> adj[100000];
int color[100000];
int low[100000];
int high[100000];
int lowColor[100000];
int cCnt[100000];
int kCnt[100001];
int timer;
int sr;
void dfs(int u, int p = ~0) {
  low[u] = ++timer;
  lowColor[timer] = color[u];
  for (int &v : adj[u])
    if (v != p) dfs(v, u);
  high[u] = timer;
}
const bool query::operator<(const query &rhs) const {
  if (this->l / sr != rhs.l / sr) return this->l / sr < rhs.l / sr;
  return this->r < rhs.r;
}
void addColor(int color) { ++kCnt[++cCnt[color]]; }
void removeColor(int color) { --kCnt[cCnt[color]--]; }
inline void run() {
  int ans[100000];
  int n, m, L = 0, R = ~0;
  scanf("%d%d", &n, &m);
  timer = ~0;
  sr = (int)sqrtl(n) + 1;
  for (int i = 0; i < n;) scanf("%d", color + i++);
  for (int i = 0; i < n;) --color[i++];
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[--u].emplace_back(--v);
    adj[v].emplace_back(u);
  }
  dfs(0);
  for (int i = 0; i < m; ++i) {
    int u;
    scanf("%d%d", &u, &q[i].k);
    q[i].l = low[--u];
    q[i].r = high[u];
    q[i].ind = i;
  }
  sort(q, q + m);
  for (int i = 0; i < m; ++i) {
    while (L < q[i].l) removeColor(lowColor[L++]);
    while (R > q[i].r) removeColor(lowColor[R--]);
    while (L > q[i].l) addColor(lowColor[--L]);
    while (R < q[i].r) addColor(lowColor[++R]);
    ans[q[i].ind] = kCnt[q[i].k];
  }
  for (int i = 0; i < m;) printf("%d\n", ans[i++]);
  for (int i = 0; i < n;) adj[i++].clear();
  memset(cCnt, 0, sizeof cCnt);
  memset(kCnt, 0, sizeof kCnt);
}
int main() {
  FILE *input = stdin;
  FILE *output = stdout;
  while (!feof(input)) {
    run();
    break;
  }
  return 0;
}