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 = 1e5 + 10;
vector<pair<int, int> > q[N];
vector<int> v[N];
int n, m;
int ans[N], a[N];
int x[N << 2];
void modify(int pos, int val, int st = 1, int en = n, int rt = 1) {
  x[rt] += val;
  if (st == en) return;
  int md = (st + en) >> 1;
  if (pos <= md)
    modify(pos, val, st, md, rt << 1);
  else
    modify(pos, val, md + 1, en, rt << 1 | 1);
}
int query(int l, int r, int st = 1, int en = n, int rt = 1) {
  if (l > en || r < st) return 0;
  if (l <= st && en <= r) return x[rt];
  int md = (st + en) >> 1;
  return query(l, r, st, md, rt << 1) + query(l, r, md + 1, en, rt << 1 | 1);
}
int main(void) {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d%d", &l, &r);
    q[r].push_back({l, i});
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] > n) continue;
    v[a[i]].push_back(i);
    int sz = v[a[i]].size();
    if (sz >= a[i]) modify(v[a[i]][sz - a[i]], 1);
    if (sz >= a[i] + 1) modify(v[a[i]][sz - a[i] - 1], -2);
    if (sz >= a[i] + 2) modify(v[a[i]][sz - a[i] - 2], 1);
    for (auto& p : q[i]) {
      ans[p.second] = query(p.first, i);
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
}
