The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai. 

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

Input

7 2
3 1 2 2 3 3 7
1 7
3 4


Output

3
1
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 300;
const long long Mod = 998244353, OO = 2e9;
int root, A[N];
unordered_map<int, int> freq;
unordered_map<int, bool> ok1, ok2;
vector<int> Sol(N);
struct query {
  int l, r, idx;
  query() {}
  bool operator<(const query &qu) const {
    if (l / root != qu.l / root) {
      return l < qu.l;
    }
    return r < qu.r;
  }
};
int ans = 0;
void add(int idx) {
  freq[A[idx]]++;
  if (freq[A[idx]] == A[idx]) {
    ++ans;
    ok1[A[idx]] = true;
  }
  if (freq[A[idx]] > A[idx] && ok1[A[idx]]) {
    --ans;
    ok1[A[idx]] = false;
  }
}
void remove(int idx) {
  freq[A[idx]]--;
  if (freq[A[idx]] == A[idx]) {
    ++ans;
    ok1[A[idx]] = true;
  }
  if (freq[A[idx]] < A[idx] && ok1[A[idx]]) {
    --ans;
    ok1[A[idx]] = false;
  }
}
query qu[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  root = sqrt(n) + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", A + i);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &qu[i].l, &qu[i].r);
    qu[i].idx = i;
  }
  sort(qu + 1, qu + 1 + m);
  int l = 1, r = 0;
  for (int i = 1; i <= m; ++i) {
    while (r < qu[i].r) {
      r++;
      add(r);
    }
    while (r > qu[i].r) {
      remove(r);
      --r;
    }
    while (l < qu[i].l) {
      remove(l);
      ++l;
    }
    while (l > qu[i].l) {
      --l;
      add(l);
    }
    Sol[qu[i].idx] = ans;
  }
  for (int i = 1; i <= m; ++i) {
    printf("%d ", Sol[i]);
  }
  printf("\n");
}
