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 Thread(null, new Solution(), "", 1<<26).start();
	}

}