You are given an array a_1, a_2, …, a_n of integers. This array is non-increasing.

Let's consider a line with n shops. The shops are numbered with integers from 1 to n from left to right. The cost of a meal in the i-th shop is equal to a_i.

You should process q queries of two types:

  * 1 x y: for each shop 1 ≤ i ≤ x set a_{i} = max(a_{i}, y). 
  * 2 x y: let's consider a hungry man with y money. He visits the shops from x-th shop to n-th and if he can buy a meal in the current shop he buys one item of it. Find how many meals he will purchase. The man can buy a meal in the shop i if he has at least a_i money, and after it his money decreases by a_i. 

Input

The first line contains two integers n, q (1 ≤ n, q ≤ 2 ⋅ 10^5).

The second line contains n integers a_{1},a_{2}, …, a_{n} (1 ≤ a_{i} ≤ 10^9) — the costs of the meals. It is guaranteed, that a_1 ≥ a_2 ≥ … ≥ a_n.

Each of the next q lines contains three integers t, x, y (1 ≤ t ≤ 2, 1≤ x ≤ n, 1 ≤ y ≤ 10^9), each describing the next query.

It is guaranteed that there exists at least one query of type 2.

Output

For each query of type 2 output the answer on the new line.

Example

Input


10 6
10 10 10 6 6 5 5 5 3 1
2 3 50
2 4 10
1 3 10
2 2 36
1 4 7
2 2 17


Output


8
3
6
2

Note

In the first query a hungry man will buy meals in all shops from 3 to 10.

In the second query a hungry man will buy meals in shops 4, 9, and 10.

After the third query the array a_1, a_2, …, a_n of costs won't change and will be \{10, 10, 10, 6, 6, 5, 5, 5, 3, 1\}.

In the fourth query a hungry man will buy meals in shops 2, 3, 4, 5, 9, and 10.

After the fifth query the array a of costs will be \{10, 10, 10, 7, 6, 5, 5, 5, 3, 1\}.

In the sixth query a hungry man will buy meals in shops 2 and 4.
1 1
5
2 1 6

1

###rangesum query using binary search and pointupdate(greedy)
n,q=[int(x) for x in input().split()]
arr=[int(t) for t in input().split()]
rangesum=[None for x in range(n)]
rangesum[-1]=arr[-1]
for i in reversed(range(n-1)):
    rangesum[i]=rangesum[i+1]+arr[i]
def search(start,end,val):###first value from left less than of equal to val
    if start==end:
        if arr[start]==val:
            ans=start
        elif arr[start]<val:
            ans= start
        else:
            ans=False
        return ans

    mid=(start+end)//2
    if arr[mid]==val:
        if mid==start:
            return mid
        elif arr[mid-1]==val:
            ans=search(start,mid-1,val)
        else:
            return mid

    elif arr[mid]>val:
        ans=search(mid+1,end,val)
    else:
        if start==mid:
            return mid
        elif arr[mid-1]>val:
            return mid
        else:
            ans=search(start,mid-1,val)
    return ans
for i in range(q):
    t,x,y=[int(x) for x in input().split()]
    if t==2:
        ans=0
        moneyleft=True;shopleft=True
        while moneyleft and shopleft:
            index=search(x-1,n-1,y)
            if index!=False:
                y-=arr[index]
                ans+=1
                x+=1
                if x>n:shopleft=False
                if y<=0:moneyleft=False
            else:
                shopleft=False
        print(ans)
    if t==1:
        index=x-1
        while index>=0:
            if arr[index]<y:
                arr[index]=y
                index-=1
            else:
                break
