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;
int n, q, m, a[1000005], s[1000005], ans[1000005], res, b[1000005];
struct node {
  int l, r, d;
} c[1000005];
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x * f;
}
inline int cmp(node x, node y) {
  return b[x.l] != b[y.l] ? x.l < y.l : (b[x.l] & 1 ? x.r < y.r : x.r > y.r);
}
inline void add(register int x) {
  if (x > n) return;
  if (s[x] == x) res--;
  s[x]++;
  if (s[x] == x) res++;
}
inline void del(register int x) {
  if (x > n) return;
  if (s[x] == x) res--;
  s[x]--;
  if (s[x] == x) res++;
}
int main() {
  n = read();
  q = read();
  m = sqrt(q);
  for (register int i = 1; i <= n; ++i) a[i] = read(), b[i] = (i - 1) / m + 1;
  for (register int i = 1; i <= q; ++i)
    c[i].l = read(), c[i].r = read(), c[i].d = i;
  sort(c + 1, c + 1 + q, cmp);
  int l = 0, r = 0;
  for (register int i = 1; i <= q; ++i) {
    while (r < c[i].r) add(a[++r]);
    while (r > c[i].r) del(a[r--]);
    while (l < c[i].l) del(a[l++]);
    while (l > c[i].l) add(a[--l]);
    ans[c[i].d] = res;
  }
  for (register int i = 1; i <= q; ++i) printf("%d\n", ans[i] + 1);
  return 0;
}
