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
BLOCK = 316
n,q = map(int,input().split())
arr = [int(x) for x in input().split()]
query = []
# storing the queries ans
ans = [0] * q
# storing their frequency inside the curr block
freq = [0] * (n+1)
# window means the numbers that have freq[x] = x
window = 0

def operate(val, n, what):
    global window
    if(val > n):
        return -1
    if(what < 0):
        if(freq[val] == val):
            window -= 1
        freq[val]-=1
        if(freq[val] == val):
            window += 1
    else:
        if(freq[val] == val):
            window -= 1
        freq[val]+=1
        if(freq[val] == val):
            window += 1

for i in range(q):
    a,b = map(int,input().split())
    query.append([a-1,b-1,i])
query.sort(key = lambda x: (x[0]//BLOCK, x[1]))
L = 0
R = -1
for l,r,idx in query:
    while(R < r):
        R += 1
        operate(arr[R], n, 1)
    while(L > l):
        L -= 1
        operate(arr[L], n, 1)
    while(R > r):
        operate(arr[R], n, -1)
        R -= 1
    while(L < l):
        operate(arr[L], n, -1)
        L += 1
    ans[idx] = window
for i in ans:
    print(i)