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;
pair<pair<int, int>, int> consultas[(100010)];
int memo[(100010)];
int v[(100010)];
int N, M;
bool compara(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second != y.first.second)
    return (x.first.second < y.first.second);
  return x < y;
}
int bit[(100010)];
int get(int p) {
  int r = 0;
  while (p > 0) {
    r += bit[p];
    p -= (p & (-p));
  }
  return r;
}
void update(int x, int p) {
  while (p <= N) {
    bit[p] += x;
    p += (p & (-p));
  }
}
vector<int> pos[(100010)];
void anda(int p) {
  int x = v[p];
  if (x > N) return;
  pos[x].push_back(p);
  int t = (int)pos[x].size();
  if (t < x) return;
  if (t == x) {
    update(1, pos[x][0]);
    return;
  }
  if (t == x + 1) {
    update(-2, pos[x][0]);
    update(+1, pos[x][1]);
    return;
  }
  update(+1, pos[x][t - x - 2]);
  update(-1, pos[x][t - x - 1]);
  update(-1, pos[x][t - x - 1]);
  update(+1, pos[x][t - x]);
}
int main() {
  scanf(" %d %d", &N, &M);
  for (int i = 1; i <= N; i++) {
    scanf(" %d", v + i);
    pos[i].clear();
  }
  for (int i = 0; i < M; i++) {
    scanf("%d %d", &consultas[i].first.first, &consultas[i].first.second);
    consultas[i].second = i;
  }
  sort(consultas, consultas + M, compara);
  memset(bit, 0, sizeof bit);
  int right = 0;
  for (int i = 0; i < M; i++) {
    int a = consultas[i].first.first;
    int b = consultas[i].first.second;
    while (right < b) {
      right++;
      anda(right);
    }
    memo[consultas[i].second] = get(b) - get(a - 1);
  }
  for (int i = 0; i < M; i++) printf("%d\n", memo[i]);
  return 0;
}
