Problem:
Score : 400 points

Problem Statement

Given is a rooted tree with N vertices numbered 1 to N.
The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i.

Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0.

Now, the following Q operations will be performed:

Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j.

Find the value of the counter on each vertex after all operations.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq Q \leq 2 \times 10^5

1 \leq a_i < b_i \leq N

1 \leq p_j \leq N

1 \leq x_j \leq 10^4

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N Q
a_1 b_1
:
a_{N-1} b_{N-1}
p_1 x_1
:
p_Q x_Q

Output

Print the values of the counters on Vertex 1, 2, \ldots, N after all operations, in this order, with spaces in between.

Sample Input 1

4 3
1 2
2 3
2 4
2 10
1 100
3 1

Sample Output 1

100 110 111 110

The tree in this input is as follows:

Each operation changes the values of the counters on the vertices as follows:

Operation 1: Increment by 10 the counter on every vertex contained in the subtree rooted at Vertex 2, that is, Vertex 2, 3, 4. The values of the counters on Vertex 1, 2, 3, 4 are now 0, 10, 10, 10, respectively.

Operation 2: Increment by 100 the counter on every vertex contained in the subtree rooted at Vertex 1, that is, Vertex 1, 2, 3, 4. The values of the counters on Vertex 1, 2, 3, 4 are now 100, 110, 110, 110, respectively.

Operation 3: Increment by 1 the counter on every vertex contained in the subtree rooted at Vertex 3, that is, Vertex 3. The values of the counters on Vertex 1, 2, 3, 4 are now 100, 110, 111, 110, respectively.

Sample Input 2

6 2
1 2
1 3
2 4
3 6
2 5
1 10
1 10

Sample Output 2

20 20 20 20 20 20

Solution in Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

module pair_stack
    implicit none
    private
    integer, allocatable:: stack(:)
    integer:: depth, max_depth
    public:: init, pop, push, is_remaining

contains
    subroutine init(x)
        integer, intent(in):: x
        max_depth = x
        allocate(stack(x))
        stack(:) = 0
        depth = 0
    end subroutine

    subroutine pop(a)
        integer, intent(out):: a
        if (depth <= 0)stop '-- --Stack is empty'
        a = stack(depth)
        depth = depth - 1
        ! print*, '-- -- poped =>', a, b
    end subroutine

    subroutine push(a)
        integer, intent(in):: a
        if (depth >= max_depth)stop '-- --Stack over flow'
        ! print*, '-- -- pushed =>', a, b
        depth = depth + 1
        stack(depth) = a
    end subroutine

    function is_remaining() result(ret)
        logical:: ret
        if (depth <= 0)then
            ret = .false.
        else
            ret = .true.
        end if
    end function
end module


program main
    implicit none
    type to
        integer,allocatable :: arr(:)
    end type
    integer:: N, Q
    integer, allocatable:: px(:, :), counter(:), ab(:, :), cnum(:)
    type(to),allocatable :: tree(:)
    integer:: i,a,b

    read*, N,Q
    allocate(px(2, Q), counter(N), ab(2, N-1), cnum(N), source=0)
    allocate(tree(N))


    do i=1, N-1
        read*, a,b
        cnum(a) = cnum(a) + 1
        cnum(b) = cnum(b) + 1
        ab(1,i) = a
        ab(2,i) = b
    enddo

    read*, px(1:2, 1:Q)

    do i=1,N
        allocate(tree(i)%arr(cnum(i)),source=0)
    enddo

    cnum(1:N) = 1

    do i=1,N-1
        a = ab(1, i)
        b = ab(2, i)
        tree(a)%arr(cnum(a)) = b
        tree(b)%arr(cnum(b)) = a
        cnum(a) = cnum(a) + 1
        cnum(b) = cnum(b) + 1
    enddo



    do i=1,Q
        counter(px(1, i)) = counter(px(1, i)) + px(2, i)
    enddo


    call dfs()
    
    do i=1, N
        print*, counter(i)
    enddo

contains
    subroutine dfs()
        use pair_stack
        integer:: v, u, p, tmp, child
        logical:: visited(N)
        call init(10 + int(log10(real(N))))
        call push(1)
        child = 0
        p = 0
        visited(1:N) = .false.
        do while(is_remaining())
            call pop(v)
            visited(v) = .true.
            do i=1, size(tree(v)%arr)
                u = tree(v)%arr(i)
                if(visited(u)) cycle
                call push(u)
                counter(u) = counter(u) + counter(v)
            enddo
        enddo

    end subroutine
end program main