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
from functools import lru_cache
from sys import stdin, stdout
import sys
from math import *

# from collections import deque
sys.setrecursionlimit(int(2e5+10))
# input = stdin.readline
# print = stdout.write
# dp=[-1]*100000

# lru_cache(None)

n,m=map(int,input().split())
ar=list(map(int,input().split()))
ar.insert(0,0)
ranges=[[0,0]]
for i in range(m):
	l,r=map(int,input().split())
	ranges.append([l,r])
cnt=[0]*(n+1)
for i in range(1,n+1):
	if(ar[i]<=n):
		cnt[ar[i]]+=1
ans=[0]*(m+1)
s=[0]*(n+1)
# print(cnt)
for i in range(1,n+1):
	if(cnt[i]>=i):
		for j in range(1,n+1):
			s[j]=s[j-1]
			if(ar[j]==i):
				s[j]+=1
		for j in range(1,m+1):
			if(s[ranges[j][1]]-s[ranges[j][0]-1]==i):
				ans[j]+=1
for i in range(1,m+1):
	print(ans[i])

