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;
bool debug = false;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
vector<pair<int, int>> q[100005];
int T[270000], M;
void update(int p, int v) {
  int l = p + M;
  while (l) {
    T[l] += v;
    l >>= 1;
  }
}
int qq(int l, int r) {
  int aa = 0;
  for (l = l + M - 1, r = r + M + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) aa += T[l ^ 1];
    if (r & 1) aa += T[r ^ 1];
  }
  return aa;
}
int a[100005], ans[100005];
vector<int> p[100005];
int main() {
  scanf("%d%d", &n, &m);
  for (M = 1; n + 2 > M; M <<= 1)
    ;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  int l, r;
  for (int(i) = 0; (i) < (int)(m); (i)++) {
    scanf("%d%d", &l, &r);
    q[r].push_back({l, i});
  }
  for (int i = 1; i <= n; i++) {
    int t = a[i];
    if (t <= 100000) {
      p[t].push_back(i);
      int ss = p[t].size();
      if (ss >= t) {
        update(p[t][ss - t], 1);
      }
      if (ss > t) {
        update(p[t][ss - t - 1], -2);
      }
      if (ss > t + 1) {
        update(p[t][ss - t - 2], 1);
      }
    }
    for (auto x : q[i]) {
      int l = x.first, id = x.second;
      ans[id] = qq(l, i);
    }
  }
  for (int(i) = 0; (i) < (int)(m); (i)++) printf("%d\n", ans[i]);
  return 0;
}
