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>
#pragma warning(disable : 4996)
using namespace std;
int Z;
struct Q {
  int l, r, ind;
  bool operator<(const Q &t) const {
    if (l / Z != t.l / Z) return l / Z < t.l / Z;
    return r < t.r;
  }
} q[100001];
int ans[100001], arr[100001], freq[100001];
bool check(int x) { return x <= 100000; }
int sum;
void checkF(int x) {
  if (freq[arr[x]] == arr[x]) ++sum;
  if (freq[arr[x]] == arr[x] + 1) --sum;
}
void checkS(int x) {
  if (freq[arr[x]] == arr[x]) ++sum;
  if (freq[arr[x]] == arr[x] - 1) --sum;
}
int main() {
  int n, m, l, r;
  scanf("%d%d", &n, &m);
  Z = sqrt(n) + 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].l--, q[i].r--;
    q[i].ind = i;
  }
  sort(q, q + m);
  int curL = 0, curR = -1;
  for (int i = 0; i < m; i++) {
    while (curR < q[i].r) {
      ++curR;
      if (check(arr[curR])) {
        ++freq[arr[curR]];
        checkF(curR);
      }
    }
    while (curR > q[i].r) {
      if (check(arr[curR])) {
        --freq[arr[curR]];
        checkS(curR);
      }
      --curR;
    }
    while (curL > q[i].l) {
      --curL;
      if (check(arr[curL])) {
        ++freq[arr[curL]];
        checkF(curL);
      }
    }
    while (curL < q[i].l) {
      if (check(arr[curL])) {
        --freq[arr[curL]];
        checkS(curL);
      }
      ++curL;
    }
    ans[q[i].ind] = sum;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
