human
Solve in Python:
Let $a_1, \ldots, a_n$ 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 $q$ ($1 \le n, q \le 3 \cdot 10^5$)  — the length of the array and the number of queries.

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

The $i$-th of the next $q$ lines contains two integers $x$ and $y$ ($x, y \ge 0$ 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 $[\times, \times, \times, 9, 5, 4, 6, 5, 7, 8, 3, 11, \times]$ (we represent $14$ as $\times$ for clarity).

Here is a strategy that removes $5$ elements (the element removed is colored in red):  $[\times, \times, \times, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, \times]$  $[\times, \times, \times, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, \times]$  $[\times, \times, \times, 9, 4, \color{red}{6}, 5, 7, 8, 3, \times]$  $[\times, \times, \times, 9, 4, 5, 7, \color{red}{8}, 3, \times]$  $[\times, \times, \times, 9, 4, 5, \color{red}{7}, 3, \times]$  $[\times, \times, \times, 9, 4, 5, 3, \times]$ (final state) 

It is impossible to remove more than $5$ elements, hence the weight is $5$.
assistant
#Here is Python program to print $q$ lines, $i$-th line should contain a single integer  — the answer to the $i$-th query.

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 = list(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 = list(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)