Problem:
Score : 300 points

Problem Statement

We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.

It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).

Consider the following action:

Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.

How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Constraints

1 \leq N, M \leq 200000

1 \leq K \leq 10^9

1 \leq A_i, B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M K
A_1 A_2 \ldots A_N
B_1 B_2 \ldots B_M

Output

Print an integer representing the maximum number of books that can be read.

Sample Input 1

3 4 240
60 90 120
80 150 80 150

Sample Output 1

3

In this case, it takes us 60, 90, 120 minutes to read the 1-st, 2-nd, 3-rd books from the top on Desk A, and 80, 150, 80, 150 minutes to read the 1-st, 2-nd, 3-rd, 4-th books from the top on Desk B, respectively.

We can read three books in 230 minutes, as shown below, and this is the maximum number of books we can read within 240 minutes.

Read the topmost book on Desk A in 60 minutes, and remove that book from the desk.

Read the topmost book on Desk B in 80 minutes, and remove that book from the desk.

Read the topmost book on Desk A in 90 minutes, and remove that book from the desk.

Sample Input 2

3 4 730
60 90 120
80 150 80 150

Sample Output 2

7

Sample Input 3

5 4 1
1000000000 1000000000 1000000000 1000000000 1000000000
1000000000 1000000000 1000000000 1000000000

Sample Output 3

0

Watch out for integer overflows.

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

      MODULE bSearchMod
! 二分探索
! 昇順の配列のみ対応。
      IMPLICIT NONE
      
      integer, parameter, private :: p = 8
      
      contains
! key以上の値が現れる最初のインデックスposを返す。
! O( log2(n) )
! 配列arrayはインデックスが1スタートで昇順の状態にしておいてほしい。
! key値がarrayの最大値より大きい場合、配列サイズ+1が返される。
! ex)
! [2, 2, 4, 4, 5]; key = -1
!  pos = 1
!
! [2, 2, 4, 4, 5]; key = 3
!        pos = 3
!
! [2, 2, 4, 4, 5]; key = 100
!                 pos = 6
      function lowerBound( array,key ) result( pos )
        implicit none
        integer(p),intent(in) :: array(1:),key
        integer(p) :: pos
        integer(p) :: left,right,mid
        logical :: isOK
        
        left = 0
        right = size( array )+1
        
        do while( abs(right-left)>1 )
          mid = (left+right)/2 !オーバーフローは考えないよ
          
          isOK = key<=array(mid)
          if( isOK )then
            right = mid
          else
            left = mid
          end if
        end do
        
        pos = right
      end function lowerBound
      
      END MODULE bSearchMod
      
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

      PROGRAM tsundoku
      use bSearchMod
      IMPLICIT NONE
      INTEGER,PARAMETER :: p = 8
      integer(p) :: n,m,k,ans=0,time
      integer(p), allocatable :: a(:),b(:)
      integer(p) :: i,numB
      
      read*,n,m,k
      allocate( a(n),b(m) )
      read*,a
      read*,b
      
      if( n>=2 )then
        do i = 2,n
          a(i) = a(i-1) + a(i)
        end do
      end if
      if( m>=2 )then
        do i = 2,m
          b(i) = b(i-1) + b(i)
        end do
      end if
      
      do i = 1,n
        time = k-a(i)
        if( time<0 ) cycle
        
        numB = lowerBound(b,time+1) -1
        ans = max( ans,i+numB )
      end do
      
      print*,ans
      
      stop
      !debugg
      
      
      
      END PROGRAM