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;
struct query {
  int l, r, idx;
};
const int SQRT = 320;
int ans[100010], arr[100010];
int freq[100010] = {0};
int Ans = 0, n;
bool comp(const query &n1, const query &n2) {
  if (n1.l / SQRT == n2.l / SQRT) return n1.r < n2.r;
  return n1.l < n2.l;
}
inline void add(int idx) {
  if (arr[idx] > n) return;
  int newval = ++freq[arr[idx]];
  if (newval == arr[idx])
    Ans++;
  else if (newval == arr[idx] + 1)
    Ans--;
}
inline void remove(int idx) {
  if (arr[idx] > n) return;
  int newval = --freq[arr[idx]];
  if (newval == arr[idx])
    Ans++;
  else if (newval == arr[idx] - 1)
    Ans--;
}
inline int getAns() { return Ans; }
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios::sync_with_stdio(false);
  int Q;
  query q[100010];
  cin >> n >> Q;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  for (int i = 0; i < Q; ++i) {
    cin >> q[i].l >> q[i].r;
    q[i].l--;
    q[i].r--;
    q[i].idx = i;
  }
  sort(q, q + Q, comp);
  int currL = 0, currR = -1;
  for (int i = 0; i < Q; ++i) {
    int l = q[i].l, r = q[i].r;
    while (l < currL) add(--currL);
    while (currR < r) add(++currR);
    while (currL < l) remove(currL++);
    while (r < currR) remove(currR--);
    ans[q[i].idx] = getAns();
  }
  for (int i = 0; i < Q; ++i) printf("%d\n", ans[i]);
}
