题目链接：http://poj.org/problem?id=3370

Description

Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year's experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided.

Your job is to help the children and present a solution.

Input

The input contains several test cases.

The first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a₁ , ... , a_n (1 ≤ a_i ≤ 100000 ), where a_i represents the number of sweets the children get if they visit neighbour i.

The last test case is followed by two zeros.

Output

For each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of a_i sweets). If there is no solution where each child gets at least one sweet print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.

Sample Input

4 51 2 3 7 53 67 11 2 5 13 170 0

Sample Output

3 52 3 4

题目大意：n个小朋友，拜访m家，每家会得到a[i]块糖，使得得到的糖数总和是n的倍数，输出拜访谁家。（这里未要求连续）

解题思路： 与POJ 2356几乎一样，n<=m, 不用考虑是否是连续的和。由于抽屉原理，最终都会至少有两个模值相同的数在一个抽屉中，不影响结果。

也就是说这题不存在无解的情况

代码如下：

#include
      
       #include
       
        int sum[100003], pla[100003];int main(){    int i, n, pp, j, m, a;    while(scanf("%d%d", &n, &m)!=EOF)    {        if(n==0&&m==0) break;        sum[0]=0, pp=0;        memset(pla, 0, sizeof(pla));        for(i=1; i<=m; i++)        {            scanf("%d", &a);            sum[i]=(sum[i-1]+a)%n;            if(sum[i]==0)                pp=i;        }        if(pp!=0)        {            for(i=1; i