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 maxn = 1e6 + 100;
int a[maxn], b[maxn], c[maxn];
int belong[maxn], res[maxn];
struct node {
  int l, r, id;
} q[maxn];
int ans = 0;
int mp[maxn];
int cnt = 0;
bool cmp(node x, node y) {
  if (belong[x.l] != belong[y.l]) {
    return x.l < y.l;
  } else {
    if (belong[x.l] & 1) {
      return x.r < y.r;
    } else {
      return x.r > y.r;
    }
  }
}
void add(int pos) {
  if (mp[a[pos]] == c[pos]) {
    ans--;
  }
  mp[a[pos]]++;
  if (mp[a[pos]] == c[pos]) {
    ans++;
  }
}
void remove(int pos) {
  if (mp[a[pos]] == c[pos]) {
    ans--;
  }
  mp[a[pos]]--;
  if (mp[a[pos]] == c[pos]) {
    ans++;
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  int block = (int)sqrt(n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = c[i] = a[i];
    belong[i] = (i - 1) / block + 1;
  }
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;
  }
  for (int i = 1; i <= m; i++) {
    cin >> q[i].l >> q[i].r;
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  for (int l = 1, r = 0, i = 1; i <= m; i++) {
    int ql = q[i].l, qr = q[i].r;
    while (l < ql) {
      remove(l++);
    }
    while (l > ql) {
      add(--l);
    }
    while (r > qr) {
      remove(r--);
    }
    while (r < qr) {
      add(++r);
    }
    res[q[i].id] = ans;
  }
  for (int i = 1; i <= m; i++) {
    cout << res[i] << endl;
  }
}
