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;
const int N = 2e6 + 5;
int n, Q;
int pre[N], nxt[N], lst[N], cnt[N], pi[N], a[N];
struct node {
  int l, r, id;
} q[N];
int tr[N], ans[N];
void add(int x, int k) {
  while (x <= n + 1) {
    tr[x] += k;
    x += x & -x;
  }
}
int find(int x) {
  int ans = 0;
  while (x) {
    ans += tr[x];
    x -= x & -x;
  }
  return ans;
}
bool cmp(node a, node b) { return a.r < b.r; }
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] > n) continue;
    pre[i] = lst[a[i]];
    nxt[lst[a[i]]] = i;
    lst[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] > n) continue;
    cnt[a[i]]++;
    if (cnt[a[i]] == 1) lst[a[i]] = i;
    if (cnt[a[i]] == a[i]) {
      pi[i] = lst[a[i]];
    } else if (cnt[a[i]] > a[i]) {
      pi[i] = nxt[pi[pre[i]]];
    }
  }
  for (int i = 1; i <= Q; ++i) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q + 1, q + Q + 1, cmp);
  int l = 1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] <= n) {
      if (pi[i]) {
        add(pre[pi[i]] + 2, 1);
        add(pi[i] + 2, -1);
        if (pre[pi[i]]) {
          add(pre[pre[pi[i]]] + 2, -1);
          add(pre[pi[i]] + 2, 1);
        }
      }
    }
    while (q[l].r == i) {
      ans[q[l].id] = find(q[l].l + 1);
      l++;
    }
  }
  for (int i = 1; i <= Q; ++i) printf("%d\n", ans[i]);
}
