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.
from sys import stdin

def bitadd(a,w,bit):
 
    x = a
    while x <= (len(bit)-1):
        bit[x] += w
        x += x & (-1 * x)
 
def bitsum(a,bit):
 
    ret = 0
    x = a
    while x > 0:
        ret += bit[x]
        x -= x & (-1 * x)
    return ret

class RangeBIT:

    def __init__(self,N,indexed):
        self.bit1 = [0] * (N+2)
        self.bit2 = [0] * (N+2)
        self.mode = indexed

    def bitadd(self,a,w,bit):
 
        x = a
        while x <= (len(bit)-1):
            bit[x] += w
            x += x & (-1 * x)
 
    def bitsum(self,a,bit):
 
        ret = 0
        x = a
        while x > 0:
            ret += bit[x]
            x -= x & (-1 * x)
        return ret
    
    def add(self,l,r,w):

        l = l + (1-self.mode)
        r = r + (1-self.mode)
        self.bitadd(l,-1*w*l,self.bit1)
        self.bitadd(r,w*r,self.bit1)
        self.bitadd(l,w,self.bit2)
        self.bitadd(r,-1*w,self.bit2)

    def sum(self,l,r):
        l = l + (1-self.mode)
        r = r + (1-self.mode)
        ret =  self.bitsum(r,self.bit1) + r * self.bitsum(r,self.bit2)
        ret -= self.bitsum(l,self.bit1) + l * self.bitsum(l,self.bit2)

        return ret

n,q = map(int,stdin.readline().split())
a = list(map(int,stdin.readline().split()))

qs = [ [] for i in range(n+1) ]
ans = [None] * q

for loop in range(q):
    x,y = map(int,stdin.readline().split())
    l = x+1
    r = n-y
    qs[r].append((l,loop))

BIT = [0] * (n+1)

for r in range(1,n+1):

    b = r-a[r-1]

    if b >= 0:

        L = 1
        R = r+1
        while R-L != 1:
            M = (L+R)//2

            if bitsum(M,BIT) >= b:
                L = M
            else:
                R = M

        if bitsum(L,BIT) >= b:
            bitadd(1,1,BIT)
            bitadd(L+1,-1,BIT)


    for ql,qind in qs[r]:
        ans[qind] = bitsum(ql,BIT)

for i in ans:
    print (i)
