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 = 1e5 + 7;
int n, m, B, a[N], c[N], ans[N];
struct Node {
  int l, r, id;
  Node() {}
  Node(int _l, int _r, int _id) { l = _l, r = _r, id = _id; }
  bool operator<(const Node &p) const { return r < p.r; }
} Q[N];
vector<int> V, q[400];
int calc(int l, int r) {
  int ret = 0;
  for (int i = (l); i < (r + 1); ++i) {
    int x = a[i];
    ++c[x];
    ret += c[x] == V[x];
    ret -= c[x] == V[x] + 1;
  }
  for (int i = (l); i < (r + 1); ++i) c[a[i]] = 0;
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  B = sqrt(n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &a[i]);
    V.push_back(a[i]);
  }
  sort(V.begin(), V.end());
  V.erase(unique(V.begin(), V.end()), V.end());
  for (int i = (0); i < (n); ++i)
    a[i] = lower_bound(V.begin(), V.end(), a[i]) - V.begin();
  for (int i = (0); i < (m); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    Q[i] = Node(l - 1, r - 1, i);
  }
  sort(Q, Q + m);
  for (int i = (0); i < (m); ++i)
    if (Q[i].r - Q[i].l + 1 <= B) {
      ans[Q[i].id] = calc(Q[i].l, Q[i].r);
    } else {
      q[Q[i].l / B + (Q[i].l % B > 0)].push_back(i);
    }
  for (int i = (0); i < (B + 1); ++i) {
    int j = 0, k = 0, cnt = 0;
    for (j = B * i; j < n && k < ((int)(q[i]).size()); ++j) {
      ++c[a[j]];
      cnt += c[a[j]] == V[a[j]], cnt -= c[a[j]] == V[a[j]] + 1;
      while (k < ((int)(q[i]).size()) && j == Q[q[i][k]].r) {
        int tmp = cnt;
        for (int r = (Q[q[i][k]].l); r < (B * i); ++r) {
          ++c[a[r]];
          tmp += c[a[r]] == V[a[r]], tmp -= c[a[r]] == V[a[r]] + 1;
        }
        for (int r = (Q[q[i][k]].l); r < (B * i); ++r) --c[a[r]];
        ans[Q[q[i][k++]].id] = tmp;
      }
    }
    for (int r = (B * i); r < (j); ++r) c[a[r]] = 0;
  }
  for (int i = (0); i < (m); ++i) printf("%d\n", ans[i]);
  return 0;
}
