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 double eps = 0.000000000000001;
int n, m, sqn;
int A[100010];
int ans[100010];
int hs[100010];
struct qss {
  int l, r, ind;
};
qss que[100010];
int mas = 0;
void yolo(int ind, int val) {
  if (ind > n) return;
  if (hs[ind] == ind) mas--;
  hs[ind] += val;
  if (hs[ind] == ind) mas++;
}
bool cmp(qss a, qss b) {
  if (a.l / sqn == b.l / sqn) return a.r < b.r;
  return a.l < b.l;
}
int main() {
  scanf("%d%d", &n, &m);
  int i, j, k;
  for (i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &que[i].l, &que[i].r);
    que[i].ind = i;
  }
  sqn = sqrtl(n) * 1.25;
  sort(que, que + m, cmp);
  int l, r;
  l = 1;
  r = 1;
  mas = 0;
  yolo(A[1], 1);
  for (i = 0; i < m; i++) {
    int lx = que[i].l, rx = que[i].r;
    while (lx < l) {
      l--;
      yolo(A[l], 1);
    }
    while (r < rx) {
      r++;
      yolo(A[r], 1);
    }
    while (lx > l) {
      yolo(A[l], -1);
      l++;
    }
    while (r > rx) {
      yolo(A[r], -1);
      r--;
    }
    ans[que[i].ind] = mas;
  }
  for (i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
