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 maxn = 100010;
int tree[maxn << 2];
void update(int L, int R, int pos, int val, int v) {
  if (L == R) {
    tree[v] += val;
    return;
  }
  int mid = (L + R) >> 1;
  if (pos <= mid) update(L, mid, pos, val, v << 1);
  if (pos > mid) update(mid + 1, R, pos, val, v << 1 | 1);
  tree[v] = tree[v << 1] + tree[v << 1 | 1];
}
int query(int L, int R, int lt, int rt, int v) {
  if (lt <= L && rt >= R) return tree[v];
  int mid = (L + R) >> 1, ret = 0;
  if (lt <= mid) ret = query(L, mid, lt, rt, v << 1);
  if (rt > mid) ret += query(mid + 1, R, lt, rt, v << 1 | 1);
  return ret;
}
int a[maxn], cnt[maxn], ans[maxn];
vector<int> pos[maxn];
struct QU {
  int x, y, id;
  bool operator<(const QU &rhs) const { return y < rhs.y; }
} q[maxn];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    memset(tree, 0, sizeof tree);
    memset(cnt, 0, sizeof cnt);
    for (int i = 0; i < maxn; ++i) pos[i].clear();
    for (int i = 0; i < n; ++i) scanf("%d", a + i);
    for (int i = 0; i < m; ++i) {
      scanf("%d%d", &q[i].x, &q[i].y);
      q[i].id = i;
    }
    sort(q, q + m);
    for (int i = 0, j = 0; i < n; ++i) {
      if (a[i] <= n) {
        pos[a[i]].push_back(i + 1);
        ++cnt[a[i]];
        if (cnt[a[i]] >= a[i]) update(1, n, pos[a[i]][cnt[a[i]] - a[i]], 1, 1);
        if (cnt[a[i]] >= a[i] + 1)
          update(1, n, pos[a[i]][cnt[a[i]] - a[i] - 1], -2, 1);
        if (cnt[a[i]] > a[i] + 1)
          update(1, n, pos[a[i]][cnt[a[i]] - a[i] - 2], 1, 1);
      }
      while (j < m && q[j].y == i + 1) {
        ans[q[j].id] = query(1, n, q[j].x, q[j].y, 1);
        ++j;
      }
    }
    for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
  }
  return 0;
}
