1404_C. Fixed Point Removal
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.
```python
'''explained excellently in editorial'''
mod = 1000000007
eps = 10**-9


def main():
    import sys
    input = sys.stdin.readline

    class Bit:
        def __init__(self, n):
            self.size = n
            self.tree = [0] * (n + 1)

        def sum(self, i):
            s = 0
            while i > 0:
                s += self.tree[i]
                i -= i & -i
            return s

        def add(self, i, x):
            while i <= self.size:
                self.tree[i] += x
                i += i & -i
        #binary lifting
        def lower_bound(self, w):
            
            #if w <= 0:
            #    return 0
            x = 0
            k = 1 << (self.size.bit_length() - 1)
            while k:
                if x + k <= self.size and self.tree[x + k] > w:
                    w -= self.tree[x + k]
                    x += k
                k >>= 1
            return x + 1

    N, Q = map(int, input().split())
    A = list(map(int, input().split()))
    A = [i+1 - A[i] for i in range(N)]
    query = [[] for _ in range(N+1)]
    ans = [0] * Q
    for q in range(Q):
        x, y = map(int, input().split())
        l = x+1
        r = N-y
        query[r].append((l, q))

    bit = Bit(N+1)
    for r in range(1, N+1):
        a = A[r-1]
        if a == 0:
            bit.add(1, +1)
            bit.add(r+1, -1)
        elif a > 0:
            #if the value of l is greater than k, it means that we cannot
            #remove the particular number from the array, as there are not
            #sufficient removals before it.
            k = bit.lower_bound(a-1)
            bit.add(1, +1)
            bit.add(k, -1)
        for l, q in query[r]:
            ans[q] = bit.sum(l)
    print(*ans, sep="\n")


if __name__ == '__main__':
    main()

```