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
from collections import Counter
input=sys.stdin.readline
n,m=map(int,input().split())
a=list(map(int,input().split()))
cnt=Counter(a)
maybe={}
for val,c in cnt.items(): # count which may be in each [l:r] and label
  if c>=val:
    l=len(maybe)
    maybe[val]=l
# len(maybe)~sqrt(n)
pref=[[0]*len(maybe)]
for i in range(n):
  pref.append(pref[-1][:])
  if a[i] in maybe:
    pref[-1][maybe[a[i]]]+=1
for _ in  range(m):
  l,r=map(int,input().split())
  cur=0
  for val,c in maybe.items():
    if pref[r][c]-pref[l-1][c]==val:
      cur+=1
  print(cur)