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 Counter

n, m = map(int, input().split())
a = list(map(int, input().split()))
count = Counter(a)
maybe = {}
for k, v in count.items():
    if v >= k:
        i = len(maybe)
        maybe[k] = i
pref = [[0] * len(maybe)]
for i in range(n):
    pref.append(pref[-1][:])
    if a[i] in maybe:
        pref[-1][maybe[a[i]]] += 1
res = []
for _ in range(m):
    l, r = map(int, input().split())
    cur = 0
    for x, i in maybe.items():
        if pref[r][i] - pref[l - 1][i] == x:
            cur += 1
    res.append(cur)
print("\n".join(map(str, res)))