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 = 100005;
struct Query {
  int l, r, block, id;
  bool operator<(const Query _q) const {
    if (block == _q.block)
      return r < _q.r;
    else
      return block < _q.block;
  }
} q[maxn];
int a[maxn], ans[maxn];
int n, m;
int Ans;
int c[maxn];
void add(int x) {
  if (x >= maxn) return;
  if (c[x] == x) Ans--;
  c[x]++;
  if (c[x] == x) Ans++;
}
void del(int x) {
  if (x >= maxn) return;
  if (c[x] == x) Ans--;
  c[x]--;
  if (c[x] == x) Ans++;
}
void solve() {
  Ans = 0;
  memset(c, 0, sizeof(c));
  for (int j = q[0].l; j <= q[0].r; j++) add(a[j]);
  ans[q[0].id] = Ans;
  int L = q[0].l, R = q[0].r;
  for (int i = 1; i < m; i++) {
    for (int j = q[i].l; j < L; j++) add(a[j]);
    for (int j = R + 1; j <= q[i].r; j++) add(a[j]);
    for (int j = L; j < q[i].l; j++) del(a[j]);
    for (int j = q[i].r + 1; j <= R; j++) del(a[j]);
    ans[q[i].id] = Ans;
    L = q[i].l;
    R = q[i].r;
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int block_size = sqrt(m * 1.0);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &q[i].l, &q[i].r);
    q[i].block = q[i].l / block_size;
    q[i].id = i;
  }
  sort(q, q + m);
  solve();
  return 0;
}
