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 tri {
  int L, R, I;
};
int query[100010];
int n;
int ara[100010];
int mark[100010];
int sum;
bool cmp(tri a, tri b) {
  int k = sqrt(n) + 1;
  int Ba = a.L / k, Bb = b.L / k;
  if (Ba == Bb) return a.R < b.R;
  return Ba < Bb;
}
vector<tri> que;
void add(int x) {
  if (x > 100000 || x < 0) return;
  mark[x]++;
  int z = mark[x];
  if (z == x) sum++;
  if (z == x + 1) sum--;
}
void rem(int x) {
  if (x > 100000 || x < 0) return;
  mark[x]--;
  int z = mark[x];
  if (z == x) sum++;
  if (z == x - 1) sum--;
}
int main() {
  int i, j, k, a, b, c, x, y, z, m, q, t, l, r;
  scanf("%d %d", &n, &q);
  for (i = 0; i < n; i++) {
    scanf("%d", &ara[i + 1]);
  }
  for (i = 0; i < q; i++) {
    scanf("%d %d", &a, &b);
    tri p;
    p.L = a;
    p.R = b;
    p.I = i;
    que.push_back(p);
  }
  sort(que.begin(), que.end(), cmp);
  int left = 0, right = 1;
  ara[0] = -1e2;
  for (i = 0; i < q; i++) {
    l = que[i].L;
    r = que[i].R;
    while (right <= r) {
      add(ara[right]);
      right++;
    }
    while (right > r + 1) {
      right--;
      rem(ara[right]);
    }
    while (left < l) {
      rem(ara[left]);
      left++;
    }
    while (left > l) {
      left--;
      add(ara[left]);
    }
    query[que[i].I] = sum;
  }
  for (i = 0; i < q; i++) {
    printf("%d\n", query[i]);
  }
}
