/*
ID: wangzha4
LANG: C++
TASK: PKU3370
*/
// 根据抽屉原理一定有解
// remain[i]记录前i项和对inc的模
// if( 存在remain[i]==0 ) -- 直接输出1～i
// else remain[i]的取值是[1~n-1]
// 而remain[i]共有n个值 -- 由抽屉原理 -- 定存在i，j满足remain[i]==remain[j]
#include < stdio.h >
#include < stdlib.h >
#include < string .h >
#include < ctype.h >
#define llong unsigned long long
#define unint unsigned int
#define printcrlf printf( "\n" )
const int inf = 1000000 ;
const int size = 100010 ;
int inc, inn ;
int data[size] ;
int remain[size] ;
int flag[size] ;
int main()
{
#ifndef ONLINE_JUDGE
freopen( " in.txt " , " r " , stdin ) ;
freopen( " out.txt " , " w " ,stdout ) ;
#endif
while ( scanf( " %d %d " , & inc, & inn ) != EOF && ( inc || inn ) )
{
memset( flag, 0 , sizeof (flag) ) ;
remain[ 0 ] = 0 ;
for ( int i = 1 ; i <= inn; i ++ ){
scanf( " %d " , & data[i] ) ;
remain[i] = (remain[i - 1 ] + data[i]) % inc ;
}
for ( int i = 1 ; i <= inn; i ++ ){
if ( 0 == remain[i] ){
for ( int j = 1 ; j < i; j ++ ) printf( " %d " ,j ) ;
printf( " %d\n " ,i ) ;
break ;
}
else {
if ( flag[remain[i]] ){
for ( int j = flag[remain[i]] + 1 ; j < i; j ++ ) printf( " %d " ,j ) ;
printf( " %d\n " ,i ) ;
break ;
}
else {
flag[remain[i]] = i ;
}
}
}
} // while
return 0 ;
}