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 = 110000, M = 110000;
int n, m, a[N], ans[N], bl[M], col[N], sum;
struct question {
  int num, l, r;
} q[N];
bool operator<(question x, question y) {
  if (bl[x.l] == bl[y.l]) {
    if (bl[x.l] & 1)
      return x.r < y.r;
    else
      return x.r > y.r;
  } else
    return x.l < y.l;
}
char buf[1 << 21], *p1 = buf, *p2 = buf;
int p3 = -1;
inline int rd() {
  int tot = 0;
  char c =
      ((p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (!isdigit(c))
    c = ((p1 == p2) &&
                 (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
  while (isdigit(c))
    tot = (tot << 1) + (tot << 3) + (c ^ '0'),
    c = ((p1 == p2) &&
                 (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return tot;
}
inline void add(int x) {
  if (x > n) return;
  if (x == col[x]) sum--;
  col[x]++;
  if (x == col[x]) sum++;
}
inline void sub(int x) {
  if (x > n) return;
  if (x == col[x]) sum--;
  col[x]--;
  if (x == col[x]) sum++;
}
int main() {
  n = rd(), m = rd();
  int blo = sqrt(n);
  for (register int i = 1; i <= n; i++) a[i] = rd(), bl[i] = (i - 1) / blo + 1;
  for (register int i = 1; i <= m; i++) {
    int l, r;
    l = rd(), r = rd();
    q[i].l = l, q[i].r = r, q[i].num = i;
  }
  sort(q + 1, q + m + 1);
  for (register int i = q[1].l; i <= q[1].r; i++) add(a[i]);
  int l = q[1].l, r = q[1].r;
  for (register int i = 1; i <= m; i++) {
    while (l < q[i].l) sub(a[l++]);
    while (l > q[i].l) add(a[--l]);
    while (r < q[i].r) add(a[++r]);
    while (r > q[i].r) sub(a[r--]);
    ans[q[i].num] = sum;
  }
  for (register int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
