Let a_1, …, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).

The weight of a is defined as the maximum number of elements you can remove.

You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the length of the array and the number of queries.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — elements of the array.

The i-th of the next q lines contains two integers x and y (x, y ≥ 0 and x+y < n).

Output

Print q lines, i-th line should contain a single integer — the answer to the i-th query.

Examples

Input


13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12


Output


5
11
6
1
0


Input


5 2
1 4 1 2 4
0 0
1 0


Output


2
0

Note

Explanation of the first query:

After making first x = 3 and last y = 1 elements impossible to remove, a becomes [×, ×, ×, 9, 5, 4, 6, 5, 7, 8, 3, 11, ×] (we represent 14 as × for clarity).

Here is a strategy that removes 5 elements (the element removed is colored in red):

  * [×, ×, ×, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, ×] 
  * [×, ×, ×, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, ×] 
  * [×, ×, ×, 9, 4, \color{red}{6}, 5, 7, 8, 3, ×] 
  * [×, ×, ×, 9, 4, 5, 7, \color{red}{8}, 3, ×] 
  * [×, ×, ×, 9, 4, 5, \color{red}{7}, 3, ×] 
  * [×, ×, ×, 9, 4, 5, 3, ×] (final state) 



It is impossible to remove more than 5 elements, hence the weight is 5.
import sys
input = sys.stdin.readline

n,q = map(int,input().split())
a = list(map(int,input().split()))
a = [i-a[i]+1 for i in range(n)]
 
Q = [[] for _ in range(n)]
ans = [-1]*q
 
for _ in range(q):
    L,R = map(int,input().split())
    R = n-1-R
    Q[R].append((L,_))
 
bit = [0]*(n+1)
 
def update(idx,val):
    idx = n-1-idx
    idx += 1
    while idx < n+1:
        bit[idx] += val
        idx += idx&-idx
 
def getSum(idx):
    idx = n-1-idx
    idx += 1
    ans = 0
    while idx:
        ans += bit[idx]
        idx -= idx&-idx
    return ans
 
for R in range(n):
    if a[R] < 0 or getSum(0) < a[R]:
        for L,i in Q[R]:
            ans[i] = getSum(L)
 
        continue
 
    lo,hi = 0,R
    while lo+1 < hi:
        mid = (lo+hi)//2
        if getSum(mid) < a[R]:
            hi = mid
        else:
            lo = mid
 
    if getSum(hi) >= a[R]:
        update(hi,1)
    else:
        update(lo,1)
 
    for L,i in Q[R]:
        ans[i] = getSum(L)
print(*ans)
