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 = 2000002;
struct P {
  int l, r, o, id;
} q[N];
int n, m, i, a[N], blk, s[N], ans, p[N], L, R;
bool cmp(P x, P y) { return x.o != y.o ? x.o < y.o : x.r < y.r; }
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (a[i] > n) a[i] = n + 1;
  }
  blk = n / sqrt(m) + 1;
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].o = q[i].l / blk;
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  L = q[1].l;
  R = q[1].l - 1;
  for (i = 1; i <= m; i++) {
    while (R < q[i].r) {
      R++;
      ans -= s[a[R]] == a[R];
      s[a[R]]++;
      ans += s[a[R]] == a[R];
    }
    while (L > q[i].l) {
      L--;
      ans -= s[a[L]] == a[L];
      s[a[L]]++;
      ans += s[a[L]] == a[L];
    }
    while (R > q[i].r) {
      ans -= s[a[R]] == a[R];
      s[a[R]]--;
      ans += s[a[R]] == a[R];
      R--;
    }
    while (L < q[i].l) {
      ans -= s[a[L]] == a[L];
      s[a[L]]--;
      ans += s[a[L]] == a[L];
      L++;
    }
    p[q[i].id] = ans;
  }
  for (i = 1; i <= m; i++) printf("%d\n", p[i]);
}
