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 = 101000;
int t[N];
void inc(int i, int x) {
  for (; i < N; i |= i + 1) t[i] += x;
}
int sum(int i) {
  int r = 0;
  for (; i >= 0; i &= i + 1, i--) r += t[i];
  return r;
}
int sum(int i, int j) { return sum(j) - sum(i - 1); }
int a[N], s[N], n, k, m, res[N], id[N];
vector<int> pl[N];
pair<pair<int, int>, int> q[N];
int indx(int x) { return lower_bound(s, s + k, x) - s; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    s[i] = a[i];
  }
  sort(s, s + n);
  k = unique(s, s + n) - s;
  for (int i = 0; i < n; i++) pl[indx(a[i])].push_back(i);
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    q[i] = make_pair(make_pair(r, l), i);
  }
  sort(q, q + m);
  int j = 0;
  for (int i = 0; i < m; i++) {
    int l, r, I;
    l = q[i].first.second;
    r = q[i].first.first;
    I = q[i].second;
    for (; j <= r; j++) {
      int x = a[j], ix = indx(x);
      int &thatIdx = id[ix];
      if (pl[ix][thatIdx] != j) cout << "FUU" << endl;
      if (thatIdx - x - 1 >= 0) {
        inc(pl[ix][thatIdx - x - 1], 1);
      }
      if (thatIdx - x >= 0) {
        inc(pl[ix][thatIdx - x], -2);
      }
      if (thatIdx - x + 1 >= 0) {
        inc(pl[ix][thatIdx - x + 1], 1);
      }
      thatIdx++;
    }
    res[I] = sum(l, r);
  }
  for (int i = 0; i < m; i++) printf("%d\n", res[i]);
  return 0;
}
