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 = 100000;
int f[N + 1], ans[N], a[N];
int n, m, S = (int)sqrt(N), res;
struct query {
  int l, r, i;
  query(int i, int l, int r) {
    this->i = i;
    this->l = l;
    this->r = r;
  }
  bool operator<(const query& o) const {
    if (l / S != o.l / S) return l / S < o.l / S;
    return r > o.r;
  }
};
void add(int i) {
  if (a[i] > N) return;
  f[a[i]]++;
  if (f[a[i]] == a[i]) res++;
  if (f[a[i]] == a[i] + 1) res--;
}
void remove(int i) {
  if (a[i] > N) return;
  if (f[a[i]] == a[i]) res--;
  if (f[a[i]] == a[i] + 1) res++;
  f[a[i]]--;
}
int main() {
  scanf("%d%d", &n, &m);
  S = (int)sqrt(n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  vector<query> Q;
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    Q.push_back(query(i, l, r));
  }
  sort(Q.begin(), Q.end());
  int curL = 0, curR = 0;
  res = 0;
  int ans[N];
  add(0);
  for (int i = 0; i < m; i++) {
    while (curL < Q[i].l) {
      remove(curL);
      curL++;
    }
    while (curL > Q[i].l) {
      add(curL - 1);
      curL--;
    }
    while (curR < Q[i].r) {
      add(curR + 1);
      curR++;
    }
    while (curR > Q[i].r) {
      remove(curR);
      curR--;
    }
    ans[Q[i].i] = res;
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
