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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const int N = 1e6 + 5;
struct node {
  int l, r, id, bl;
} q[N];
inline int cmp(node x, node y) {
  return x.bl == y.bl ? x.r < y.r : x.bl < y.bl;
}
int a[N], cnt[N], ans[N], n, m, sum;
inline void add(int x) {
  if (x > n) return;
  sum -= (cnt[x] == x);
  ++cnt[x];
  sum += (cnt[x] == x);
}
inline void del(int x) {
  if (x > n) return;
  sum -= (cnt[x] == x);
  --cnt[x];
  sum += (cnt[x] == x);
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  const int B = sqrt(n);
  for (int i = 1; i <= m; i++) {
    q[i].l = read(), q[i].r = read();
    q[i].bl = (q[i].l - 1) / B + 1;
    q[i].id = i;
  }
  sort(q + 1, q + m + 1, cmp);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    int x = q[i].l, y = q[i].r, id = q[i].id;
    while (r < y) add(a[++r]);
    while (r > y) del(a[r--]);
    while (l < x) del(a[l++]);
    while (l > x) add(a[--l]);
    ans[id] = sum;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
