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 n, m, a, b;
int niz[100001];
int bit[100001][640];
int ids[640];
set<int> bst;
map<int, int> mst;
map<int, int> idd;
int id = 0;
inline void update(int x, int val, int idx) {
  while (x <= n) {
    bit[x][idx] += val;
    x += (x & -x);
  }
}
inline int read(int x, int idx) {
  int ret = 0;
  while (x > 0) {
    ret += bit[x][idx];
    x -= (x & -x);
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &niz[i]);
    if (mst[niz[i]] == niz[i] - 1) {
      bst.insert(niz[i]);
      idd[niz[i]] = id;
      ids[id++] = niz[i];
      mst[niz[i]]++;
    } else
      mst[niz[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (bst.find(niz[i]) != bst.end()) {
      update(i, 1, idd[niz[i]]);
    }
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    int sol = 0;
    for (int j = 0; j < id; j++) {
      if (read(b, j) - read(a - 1, j) == ids[j]) sol++;
    }
    printf("%d\n", sol);
  }
  return 0;
}
