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.
# Fast IO (be careful about bytestring, not on interactive)

import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

MaxLength = 2 ** 19
SegmentTree = []

CurLen = MaxLength
while CurLen != 0:
    SegmentTree.append([0] * CurLen)
    CurLen //= 2

Depth = len(SegmentTree)

def addElem(n,index): # add n to an element in index
    for j in range(Depth):
        if index < len(SegmentTree[j]):
            SegmentTree[j][index] += n
        index //= 2

def firstNSum(n): # Return the sum of the first n elements (0 to n-1)
    nCpy = n
    summ = 0
    depthCur = 0
    while nCpy != 0:
        if nCpy % 2 != 0:
            summ += SegmentTree[depthCur][nCpy - 1]
        nCpy //= 2
        depthCur += 1
    return summ


n,q = map(int,input().split())
a = list(map(int,input().split()))
for i in range(n):
    a[i] = i + 1 - a[i]
ans = [-1] * q
query = []
for i in range(q):
    x,y = map(int,input().split())
    query.append((i,x,n - y))
query.sort(key = lambda x: x[2])


currentY = 0
currentTotalSum = 0
for qu in query:
    while currentY < qu[2]:
        if a[currentY] >= 0 and currentTotalSum >= a[currentY]:
            l = 0
            r = currentY + 1
            while l < r:
                m = (l + r) // 2
                if a[currentY] > currentTotalSum - firstNSum(m):
                    r = m
                else:
                    l = m + 1
            addElem(1,r-1)
            currentTotalSum += 1
        currentY += 1
    ans[qu[0]] = currentTotalSum - firstNSum(qu[1])

for elem in ans:
    print(elem)

