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;
int n, m;
struct que {
  int s, e, id;
  que(int a, int b, int c) : s(a), e(b), id(c) {}
  bool operator<(const que &ano) const { return e < ano.e; }
};
struct segtree {
  int add[100005 << 2];
  void init() { memset(add, 0, sizeof(add)); }
  void pushdown(int rt) {
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    add[rt] = 0;
  }
  void update(int L, int R, int c, int l, int r, int rt) {
    if (L <= l && r <= R) {
      add[rt] += c;
      return;
    }
    pushdown(rt);
    int m = (l + r) >> 1;
    if (L <= m) update(L, R, c, l, m, rt << 1);
    if (m < R) update(L, R, c, m + 1, r, rt << 1 | 1);
  }
  int query(int pos, int l, int r, int rt) {
    if (pos == l && pos == r) return add[rt];
    pushdown(rt);
    int m = (l + r) >> 1;
    if (pos <= m) return query(pos, l, m, rt << 1);
    if (m < pos) return query(pos, m + 1, r, rt << 1 | 1);
  }
} it;
vector<que> q;
vector<int> pos[100005];
pair<int, int> pre[100005];
int c[100005];
int a[100005], ans[100005];
void doit() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  q.clear();
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    q.push_back(que(x, y, i));
  }
  sort(q.begin(), q.end());
  it.init();
  for (int i = 1; i <= n; i++) pos[i].clear();
  int ind = 0;
  for (int i = 1; i <= n; i++) {
    int now = a[i];
    if (now <= n) {
      c[now]++;
      pos[now].push_back(i);
      if (c[now] == now) {
        pre[now] = make_pair(1, pos[now][0]);
        it.update(pre[now].first, pre[now].second, 1, 1, n, 1);
      } else if (c[now] > now) {
        it.update(pre[now].first, pre[now].second, -1, 1, n, 1);
        pre[now] = make_pair(pre[now].second + 1, pos[now][c[now] - now]);
        it.update(pre[now].first, pre[now].second, 1, 1, n, 1);
      }
    }
    while (ind < m && q[ind].e == i) {
      ans[q[ind].id] = it.query(q[ind].s, 1, n, 1);
      ind++;
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
int main() {
  doit();
  return 0;
}
