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 = 2e5 + 9;
int inq1[maxn], c1[maxn], c2[maxn], c3[maxn], sum[maxn];
int ans;
struct node {
  int l, r, id, blk;
} edge[maxn];
bool cmp(node f1, node f2) {
  if (f1.blk == f2.blk) return f1.r < f2.r;
  return f1.l < f2.l;
}
void init() {
  memset(inq1, 0, sizeof(inq1));
  ans = 0;
}
void up(int x) {
  if (inq1[c1[x]] == c3[x]) ans--;
  inq1[c1[x]]++;
  if (inq1[c1[x]] == c3[x]) ans++;
}
void down(int x) {
  if (inq1[c1[x]] == c3[x]) ans--;
  inq1[c1[x]]--;
  if (inq1[c1[x]] == c3[x]) ans++;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    init();
    int block = sqrt(n * 1.0);
    for (int i = 1; i <= n; i++) scanf("%d", &c1[i]), c3[i] = c2[i] = c1[i];
    sort(c2 + 1, c2 + n + 1);
    int f = unique(c2 + 1, c2 + n + 1) - (c2 + 1);
    for (int i = 1; i <= n; i++)
      c1[i] = lower_bound(c2 + 1, c2 + f + 1, c1[i]) - (c2 + 1) + 1;
    for (int i = 0; i < m; i++) {
      scanf("%d%d", &edge[i].l, &edge[i].r);
      edge[i].id = i;
      edge[i].blk = edge[i].l / block;
    }
    sort(edge, edge + m, cmp);
    int l = 1, r = 0;
    for (int i = 0; i < m; i++) {
      int s = edge[i].l, t = edge[i].r;
      while (r < t) {
        r++;
        up(r);
      }
      while (l > s) {
        l--;
        up(l);
      }
      while (r > t) {
        down(r);
        r--;
      }
      while (l < s) {
        down(l);
        l++;
      }
      sum[edge[i].id] = ans;
    }
    for (int i = 0; i < m; i++) {
      printf("%d\n", sum[i]);
    }
  }
  return 0;
}
