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
import java.io.*; 
import java.util.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {
    
    public void solve () throws Exception {
        int n = nextInt();
        int m = nextInt();
        
        int a [] = new int [n + 1];
        int count [] = new int [n + 1];
        
        int k = 0;
        int arr [] = new int [n + 1];
        int pos [] = new int [n + 1];
        
        for (int i = 0; i <= n; i++) {
            pos [i] = -1;
        }
        
        for (int i = 1; i <= n; i++) {
            a [i] = nextInt();
            if (a[i] <= n)
                count [a[i]]++;
        }
        
        for (int i = 1; i <= n; i++) {
            if (count[i] >= i) {
                arr [k] = i;
                pos [i] = k;
                k++;
            }
        }
        
        int d [] = new int [(n + 1) * k + 1];
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < k; j++)
                d[i * k + j] += d[(i - 1) * k + j];
            
            if (a[i] <= n && pos[a[i]] != -1)
                d[i * k + pos[a[i]]]++;
        }
        
        for (int j = 0; j < m; j++) {
            int l = nextInt();
            int r = nextInt();
            int ret = 0;
            for (int i = 0; i < k; i++)
                if (d[r * k + i] - d[(l - 1) * k + i] == arr[i])
                    ret++;
            
            out.println(ret);
        }
        
    }
    
    private static final String fname = "";
    private static long stime = 0;
    
    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer st;
    
    private String nextToken () throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
    
    private int nextInt () throws Exception {
        return Integer.parseInt(nextToken());
    }
    
    private long nextLong () throws Exception {
        return Long.parseLong(nextToken());
    }
    
    private double nextDouble () throws Exception {
        return Double.parseDouble(nextToken());
    }
    
    @Override
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            solve ();
        } catch (Exception e) {
            throw new RuntimeException(e);
        } finally {
            out.close();
        }
    }
    
    public static void main(String[] args) {
        new Solution().run();
    }

}