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 sys
def get_ints(): return map(int, sys.stdin.readline().split())
n,m=map(int,input().split())
arr=list(get_ints())
arr=[0]+arr
summ=[0]*(100001)
freq=[0]*(100001)
#print(l)
l=[0]*(100001)
r=[0]*(100001)
for ii in arr:
 if ii<=n:freq[ii]+=1
for i  in range(1,m+1):
  x,y=get_ints()
  l[i]=x
  r[i]=y
ans=[0]*(m+1)
for i in range(1,n+1):
  if freq[i]>=i:
    for j in range(1,n+1):
      summ[j]=summ[j-1]
      if arr[j]==i:
        summ[j]+=1
    for k in range(1,m+1):
      if summ[r[k]]-summ[l[k]-1]==i:
        ans[k]+=1
for i in  range(1,m+1):
  print(ans[i])
