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;
struct IOS {
  struct Input {
    int readInt() {
      int x;
      scanf("%d", &x);
      return x;
    }
    long long readLong() {
      long long x;
      scanf("%lld", &x);
      return x;
    }
    char readChar() {
      char c;
      scanf("%c", &c);
      return c;
    }
    double readDouble() {
      double x;
      scanf("%lf", &x);
      return x;
    }
    string readString() {
      char str[100005];
      scanf("%s", &str);
      return str;
    }
    void readIntArray(int* arr, int st, int en) {
      for (int i = st; i <= en; i++) arr[i] = readInt();
    }
  } in;
  struct Output {
    void newLine() { printf("\n"); }
  } out;
};
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    std::cout << v << ' ';
    return *this;
  }
} dbg;
int arr[100005];
int cnt[100005];
int ans[100005];
vector<int> useable;
vector<pair<int, int> > Q;
class BIT {
 public:
  long long n, tree[100005];
  BIT() {}
  void Init(int _n) { n = _n; }
  void Reset() { memset(tree, 0, sizeof tree); }
  void Update(int idx, long long val) {
    for (; idx <= n; idx += idx & (-idx)) tree[idx] += val;
  }
  long long Query(int idx) {
    long long res = 0;
    for (; idx; idx -= idx & (-idx)) res += tree[idx];
    return res;
  }
  long long Query(int l, int r) { return Query(r) - Query(l - 1); }
};
BIT tree;
int main() {
  IOS ios;
  int n = ios.in.readInt();
  int m = ios.in.readInt();
  tree.Init(n);
  for (int i = 0; i <= n - 1; i++) {
    arr[i] = ios.in.readInt();
    if (arr[i] <= n) cnt[arr[i]]++;
  }
  for (int i = 1; i <= n; i++)
    if (cnt[i] >= i) useable.push_back(i);
  for (int i = 0; i <= m - 1; i++) {
    int l = ios.in.readInt();
    int r = ios.in.readInt();
    Q.push_back({l, r});
  }
  for (int i = 0; i < useable.size(); i++) {
    tree.Reset();
    int val = useable[i];
    for (int j = 0; j <= n - 1; j++)
      if (arr[j] == val) tree.Update(j + 1, 1);
    for (int j = 0; j <= m - 1; j++) {
      int l = Q[j].first;
      int r = Q[j].second;
      ans[j] += (tree.Query(l, r) == val);
    }
  }
  for (int i = 0; i <= m - 1; i++) printf("%d\n", ans[i]);
  return 0;
}
