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;
int a[100001];
int s;
int res = 0;
int fake[100001];
int conta[100001];
int ans[100001];
struct query {
  int l, r, id;
  query() {}
};
query q[100001];
bool comp(query a, query b) {
  if (a.l / s == b.l / s) {
    return a.r < b.r;
  } else {
    return a.l / s < b.l / s;
  }
}
map<int, int> trad;
void increase(int r0) {
  conta[fake[r0]]++;
  if (conta[fake[r0]] == a[r0]) {
    res++;
  }
  if (conta[fake[r0]] == a[r0] + 1) {
    res--;
  }
}
void decrease(int r0) {
  conta[fake[r0]]--;
  if (conta[fake[r0]] == a[r0]) {
    res++;
  }
  if (conta[fake[r0]] == a[r0] - 1) {
    res--;
  }
}
int main() {
  int n, m, c = 0;
  scanf("%d %d", &n, &m);
  s = sqrt(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (!trad.count(a[i])) {
      trad[a[i]] = c++;
    }
    fake[i] = trad[a[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].id = i;
  }
  sort(q, q + m, comp);
  int l0 = 0, r0 = -1;
  for (int i = 0; i < m; i++) {
    int l = q[i].l, r = q[i].r;
    while (l0 < l) {
      decrease(l0);
      l0++;
    }
    while (l0 > l) {
      l0--;
      increase(l0);
    }
    while (r0 < r) {
      r0++;
      increase(r0);
    }
    while (r0 > r) {
      decrease(r0);
      r0--;
    }
    ans[q[i].id] = res;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
