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 cnt[311111], a[311111], answer, ans[100005];
struct node {
  int L, R, i;
} q[311111];
bool cmp(node x, node y) {
  if (x.L / 555 != y.L / 555) {
    return x.L / 555 < y.L / 555;
  }
  return x.R < y.R;
}
void add(int position) {
  if (a[position] <= 100000) {
    if (cnt[a[position]] == a[position] - 1) {
      answer++;
    } else if (cnt[a[position]] == a[position]) {
      answer--;
    }
    cnt[a[position]]++;
  }
}
void remove(int position) {
  if (a[position] <= 100000) {
    if (cnt[a[position]] == a[position] + 1) {
      answer++;
    } else if (cnt[a[position]] == a[position]) {
      answer--;
    }
    cnt[a[position]]--;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < n; i++) scanf("%d", &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].i = i;
  }
  sort(q, q + m, cmp);
  int currentL = 0, currentR = 0;
  for (int i = 0; i < m; i++) {
    int L = q[i].L, R = q[i].R;
    while (currentL < L) {
      remove(currentL);
      currentL++;
    }
    while (currentL > L) {
      add(currentL - 1);
      currentL--;
    }
    while (currentR <= R) {
      add(currentR);
      currentR++;
    }
    while (currentR > R + 1) {
      remove(currentR - 1);
      currentR--;
    }
    ans[q[i].i] = answer;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
