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 bisect
import collections
import heapq
import math
def do():
    n, n_query = map(int, input().split(" "))
    N = max(n, n_query) + 1

    nums = [0] * N
    count = [0] * N
    tmp = [int(c) for c in input().split(" ")]
    for i,d in enumerate(tmp):
        nums[i+1] = d
        if d <= n:
            count[d] += 1
    l = [0] * N
    r = [0] * N
    s = [0] * N
    res = [0] * N
    for i in range(1, n_query + 1):
        l[i], r[i] = map(int, input().split(" "))
    for i in range(1, n + 1):
        if i <= count[i]:
            for j in range(1, n + 1):
                s[j] = s[j - 1] + (nums[j] == i)
            for j in range(1, n_query + 1):
                if s[r[j]] - s[l[j] - 1] == i:
                    res[j] += 1
    for i in range(1, n_query + 1):
        print(res[i])

do()





