分糖果

Description

There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from1 to n. The i-th child wants to get at least a_i candies.

Jzzhu asks children to line up. Initially, the i-th child stands at thei-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:

1. Give m candies to the first child of the line.
2. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home.
3. Repeat the first two steps while the line is not empty.

Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?

Input

The first line contains two integers n, m(1 ≤ n ≤ 100; 1 ≤ m ≤ 100). The second line containsn integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 100).

Output

Output a single integer, representing the number of the last child.

Sample Input

Input

5 21 3 1 4 2

Output

4

Input

6 41 1 2 2 3 3

Output

6

这道题的大体意思是jzzhu要给小朋友们分糖，但是每一次分给每个小朋友的糖都是有限并且一样的，但是每个小朋友对糖的需求不一样，如果没有达到需求的话，就排在队尾继续等待分糖，问最后一个离开的小朋友最开始的编号是什么？

#include<stdio.h>int main(){    int n,m,i,u;    int a[101],s[101];    while(~scanf("%d%d",&n,&m))    {        for(i =0; i<n; i++)        {            a[i] = i+1;            scanf("%d",&s[i]);        }        i=0;        u=n;        while(n>1)        {            if(a[i]!=0)            {                s[i]-=m;                if(s[i]<=0)                {                    n--;                     a[i] = 0;                }            }            i++;            if(i==u&&n>1) i = 0;        }        for(i=0; i<u; i++)            if( a[i]!=0)                printf("%d\n", a[i]);    }    return 0;}