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;
queue<int> fila[100000 + 10];
int bit[100000 + 10];
int rsp[100000 + 10];
pair<pair<int, int>, int> q[100000 + 10];
int v[100000 + 10];
int ls[100000 + 10];
void add(int p, int v) {
  if (p == 0) return;
  for (; p <= 100000; p += (p & (-p))) bit[p] += v;
}
int get(int p) {
  int ret = 0;
  for (; p > 0; p -= (p & (-p))) ret += bit[p];
  return ret;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  memset(bit, 0, sizeof(bit));
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]), ls[i] = -1, fila[i].push(0);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &q[i - 1].first.second, &q[i - 1].first.first);
    q[i - 1].second = i - 1;
  }
  sort(q, q + m);
  int ind = 0;
  for (int i = 1; i <= n; i++) {
    if (v[i] <= n) {
      if (ls[v[i]] != -1) {
        add(ls[v[i]] + 1, -1);
        add(fila[v[i]].front() + 1, +1);
      }
      if (fila[v[i]].size() == v[i]) {
        add(fila[v[i]].front() + 1, +1);
        ls[v[i]] = fila[v[i]].front();
        fila[v[i]].pop();
        fila[v[i]].push(i);
        add(fila[v[i]].front() + 1, -1);
      } else
        fila[v[i]].push(i);
    }
    while (ind < m && q[ind].first.first == i) {
      rsp[q[ind].second] = get(q[ind].first.second);
      ind++;
    }
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", rsp[i]);
  }
  return 0;
}
