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 + 5;
struct Query {
  int st, ed, id;
  Query() {}
  Query(int a, int b, int c) {
    st = a;
    ed = b;
    id = c;
  }
  bool operator<(const Query &b) const { return ed < b.ed; }
};
struct node {
  int lft, rht, sum;
  int mid() { return lft + (rht - lft) / 2; }
};
struct Segtree {
  node tree[N * 4];
  void relax(int ind) {
    int &sum = tree[ind].sum;
    if (sum != 0) {
      tree[(ind << 1)].sum += sum;
      tree[(ind << 1 | 1)].sum += sum;
      sum = 0;
    }
  }
  void build(int lft, int rht, int ind) {
    tree[ind].lft = lft;
    tree[ind].rht = rht;
    tree[ind].sum = 0;
    if (lft != rht) {
      int mid = tree[ind].mid();
      build(lft, mid, (ind << 1));
      build(mid + 1, rht, (ind << 1 | 1));
    }
  }
  void updata(int be, int end, int ind, int valu) {
    int lft = tree[ind].lft, rht = tree[ind].rht;
    if (be <= lft && rht <= end)
      tree[ind].sum += valu;
    else {
      relax(ind);
      int mid = tree[ind].mid();
      if (be <= mid) updata(be, end, (ind << 1), valu);
      if (end > mid) updata(be, end, (ind << 1 | 1), valu);
    }
  }
  int query(int pos, int ind) {
    if (tree[ind].lft == tree[ind].rht)
      return tree[ind].sum;
    else {
      relax(ind);
      int mid = tree[ind].mid();
      if (pos <= mid)
        return query(pos, (ind << 1));
      else
        return query(pos, (ind << 1 | 1));
    }
  }
} seg;
vector<Query> query;
vector<int> pos[N];
int data[N], cnt[N], res[N];
pair<int, int> pre[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &data[i]);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    query.push_back(Query(a, b, i));
  }
  sort(query.begin(), query.end());
  int ind = 0;
  seg.build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    int valu = data[i];
    if (valu <= n) {
      cnt[valu]++;
      pos[valu].push_back(i);
      if (cnt[valu] == valu) {
        pre[valu] = make_pair(1, pos[valu][0]);
        seg.updata(pre[valu].first, pre[valu].second, 1, 1);
      } else if (cnt[valu] > valu) {
        seg.updata(pre[valu].first, pre[valu].second, 1, -1);
        pre[valu] =
            make_pair(pre[valu].second + 1, pos[valu][cnt[valu] - valu]);
        seg.updata(pre[valu].first, pre[valu].second, 1, 1);
      }
    }
    while (query[ind].ed == i && ind < m) {
      res[query[ind].id] = seg.query(query[ind].st, 1);
      ind++;
    }
  }
  for (int i = 0; i < m; i++) printf("%d\n", res[i]);
  return 0;
}
