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 = 100010;
int ans[N], num[N], L, R, cnt, M[N];
struct query {
  int l, r, idx, block;
  bool operator<(const query x) const {
    if (x.block == block) return r < x.r;
    return block < x.block;
  }
} q[N];
void add(int x) {
  if (x > 100000) return;
  M[x]++;
  int occur = M[x];
  if (occur == x)
    cnt++;
  else if (occur == x + 1)
    cnt--;
}
void del(int x) {
  if (x > 100000) return;
  M[x]--;
  int occur = M[x];
  if (occur == x)
    cnt++;
  else if (occur == x - 1)
    cnt--;
}
int main() {
  int n, nq;
  cin >> n >> nq;
  int BLOCK_SIZE = (int)sqrt(n);
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  for (int i = 1; i <= nq; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].idx = i, q[i].block = q[i].l / BLOCK_SIZE;
  }
  sort(q + 1, q + nq + 1);
  for (int i = q[1].l; i <= q[1].r; i++) add(num[i]);
  ans[q[1].idx] = cnt;
  L = q[1].l, R = q[1].r;
  for (int i = 2; i <= nq; i++) {
    for (int j = L; j <= q[i].l - 1; j++) del(num[j]);
    for (int j = L - 1; j >= q[i].l; j--) add(num[j]);
    for (int j = R + 1; j <= q[i].r; j++) add(num[j]);
    for (int j = R; j >= q[i].r + 1; j--) del(num[j]);
    L = q[i].l, R = q[i].r;
    ans[q[i].idx] = cnt;
  }
  for (int i = 1; i <= nq; i++) printf("%d\n", ans[i]);
  return 0;
}
