B. Counting-out Rhyme

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本文介绍了一种通过数组模拟的算法来解决一个特定的计数淘汰游戏问题。该算法适用于小规模数据集，并能有效处理游戏中的每一步淘汰过程。

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因为n的k的数据比较小

所以可以用数组模拟

在模拟的过程中a[i]可以优化

因为是环所以a[i]=a[i]%(在场的人数

n children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to n. In the beginning, the first child is considered the leader. The game is played in k steps. In the i-th step the leader counts out ai people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.

For example, if there are children with numbers [8, 10, 13, 14, 16] currently in the circle, the leader is child 13 and ai = 12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.

You have to write a program which prints the number of the child to be eliminated on every step.

Input

The first line contains two integer numbers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ n - 1).

The next line contains k integer numbers a1, a2, ..., ak (1 ≤ ai ≤ 109).

Output

Print k numbers, the i-th one corresponds to the number of child to be eliminated at the i-th step.

Examples

input
7 5
10 4 11 4 1

output
4 2 5 6 1 

input
3 2
2 5

output
3 2 

Note

Let's consider first example:

In the first step child 4 is eliminated, child 5 becomes the leader.
In the second step child 2 is eliminated, child 3 becomes the leader.
In the third step child 5 is eliminated, child 6 becomes the leader.
In the fourth step child 6 is eliminated, child 7 becomes the leader.
In the final step child 1 is eliminated, child 3 becomes the leader.

#include<stdio.h>
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int num[103]= {0};
        int a[103];
        for(int i=0; i<m; i++)
            scanf("%d",&a[i]);
        int now=1;
        int data=0;
        for(int i=0; i<m; i++)
        {
            int yu=a[i]%(n-i);
            if(yu==0)
                yu=n-i;
            int k=0;
            for(int j=now+1; ; j++)
            {
                //printf("%d\n",k);
                if(j>n)
                    j=1;
                if(num[j]==0)
                    k++;
                if(k==yu)
                {
                    k=j;
                    break;
                }
            }
            num[k]=1;
            if(data==0)
            {
                printf("%d",k);
                data=1;
            }
            else printf(" %d",k);
            now=k;
            while(num[now])
            {
                now++;
                if(now>n)
                    now=1;
            }
        }
        printf("\n");
    }
}