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 collections import defaultdict
n,m=map(int,input().split())
query=[[] for j in range(int(n**0.5)+1)]
arr=list(map(int,input().split()))
for j in range(m):
    x,y=map(int,input().split())
    query[int(x/n**0.5)].append([x-1,y-1,j])
q=[]
for j in query:
    j.sort(key=lambda x:x[1])
    for p in j:
        q.append(p)
d=defaultdict(lambda:0)
j=0
currL,currR,x= 1,0,0
res=[0]*(m)
while(j<len(q)):
    L,R=q[j][0],q[j][1]

    while currL < L:
        d[arr[currL]]-=1
        if d[arr[currL]]==(arr[currL]-1):
            x-=1
        if d[arr[currL]] == (arr[currL]):
            x += 1
        currL += 1

    while currL > L:
        currL -= 1
        d[arr[currL]] += 1
        if d[arr[currL]]==(arr[currL]+1):
            x-=1
        if d[arr[currL]]==(arr[currL]):
            x+=1





    while currR<R:
        currR += 1
        d[arr[currR]] += 1
        if d[arr[currR]]==(arr[currR]+1):
            x-=1
        if d[arr[currR]] == (arr[currR]):
            x += 1




    while currR > R:

        d[arr[currR]] -= 1
        if d[arr[currR]]==(arr[currR]-1):
            x-=1
        if d[arr[currR]] == (arr[currR]):
            x += 1
        currR -= 1


    res[q[j][2]]=x
    j+=1
for j in res:
    print(j)

