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;
struct node {
  int l, r, idx;
};
int op[100005];
int a[100005];
node q[100005];
int sz = 316;
int n;
int ans = 0;
unordered_map<int, int> c;
bool comp(node a, node b) {
  if (((a.l) / sz) != ((b.l) / sz)) return ((a.l) / sz) < ((b.l) / sz);
  return (a.r) < (b.r);
}
void add(int i) {
  c[a[i]]++;
  if (c[a[i]] == a[i]) ans++;
  if (c[a[i]] == (a[i] + 1)) ans--;
}
void rem(int i) {
  c[a[i]]--;
  if (c[a[i]] == a[i]) ans++;
  if (c[a[i]] == (a[i] - 1)) ans--;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int left, right, i, cl, cr, m;
  cin >> n >> m;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < m; i++) {
    cin >> (q[i].l) >> (q[i].r);
    q[i].l--;
    q[i].r--;
    q[i].idx = i;
  }
  sort(q, q + m, comp);
  cl = q[0].l;
  cr = q[0].r;
  for (i = cl; i <= cr; i++) add(i);
  op[q[0].idx] = ans;
  for (i = 1; i < m; i++) {
    left = q[i].l;
    right = q[i].r;
    while (cl > left) {
      add(cl - 1);
      cl--;
    }
    while (cr < right) {
      add(cr + 1);
      cr++;
    }
    while (cl < left) {
      rem(cl);
      cl++;
    }
    while (cr > right) {
      rem(cr);
      cr--;
    }
    op[q[i].idx] = ans;
  }
  for (i = 0; i < m; i++) cout << op[i] << "\n";
  return 0;
}
