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 = 1e6 + 5;
int n, m;
int tim[maxn];
int val[maxn];
int sum;
int ans[maxn];
int pos[maxn];
int blo;
struct query {
  int l, r;
  int k;
  bool operator<(const query a) const {
    if (pos[l] == pos[a.l])
      return r < a.r;
    else
      return pos[l] < pos[a.l];
  }
} q[maxn];
template <class T>
void read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    f |= (ch == '-');
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return;
}
inline void add(int x) {
  if (x > n) return;
  if (tim[x] == x) sum--;
  tim[x]++;
  if (tim[x] == x) sum++;
}
inline void sub(int x) {
  if (x > n) return;
  if (tim[x] == x) sum--;
  tim[x]--;
  if (tim[x] == x) sum++;
}
signed main() {
  read(n), read(m);
  blo = sqrt(n);
  for (int i = 1; i <= n; i++) {
    read(val[i]);
    pos[i] = (i - 1) / blo + 1;
  }
  for (int i = 1; i <= m; i++) {
    read(q[i].l), read(q[i].r);
    q[i].k = i;
  }
  sort(q + 1, q + 1 + m);
  int l = 1, r = 0;
  for (int i = 1; i <= m; i++) {
    while (l < q[i].l) sub(val[l++]);
    while (l > q[i].l) add(val[--l]);
    while (r < q[i].r) add(val[++r]);
    while (r > q[i].r) sub(val[r--]);
    ans[q[i].k] = sum;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}
