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;
int a[100001];
typedef struct {
  int x;
  int y;
  int id;
  int r;
} query;
bool compY(query q1, query q2) { return q1.y < q2.y; }
bool compId(query q1, query q2) { return q1.id < q2.id; }
vector<query> q;
map<int, queue<int> > f;
map<int, int> p;
int bit[100001];
int add(int x, int v) {
  for (; x <= n; x += x & -x) bit[x] += v;
}
int get(int x) {
  int r = 0;
  for (; x > 0; x -= x & -x) r += bit[x];
  return r;
}
void interval(int x, int y, int v) {
  add(x, v);
  add(y + 1, -v);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < (m); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    q.push_back((query){x, y, i, 0});
  }
  sort(q.begin(), q.end(), compY);
  int j = 0;
  for (int i = 1; i <= n; i++) {
    int v = a[i];
    if (f[v].size() == v) {
      interval(p[v] + 1, f[v].front(), -1);
      p[v] = f[v].front();
      f[v].pop();
    }
    f[v].push(i);
    if (f[v].size() == v) {
      interval(p[v] + 1, f[v].front(), 1);
    }
    while (j < q.size() && i == q[j].y) {
      q[j].r = get(q[j].x);
      j++;
    }
  }
  sort(q.begin(), q.end(), compId);
  for (int i = 0; i < (q.size()); i++) {
    printf("%d\n", q[i].r);
  }
  return 0;
}
