#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

#define M 6

int d[M], tar, val, len, num[M][M];
bool flag;

struct PathCell { int l, r; } pre[M][M], ans; //记录路径
void convert(char tmp[])
{
     int i;
     for (i = 0; tmp[i] != ‘\0‘; i++)
          d[i] = tmp[i] - ‘0‘;
     len = i;
}

void preprocess()//预处理，求出剪断后的所有可能整数
{
     int i, j, k;

     for (i = 0; i < len; i++) {
         for (j = i; j < len; j++) {
             k = i;
             while (k <= j) {
                 num[i][j] = num[i][j] * 10 + d[k];
                 k++;
             }
         }
     }
}

bool dfs(int id, int now, int last)
{
    if (id >= len) {
        if (now > val && now <= tar) {
            val = now; ans.l = last; ans.r = id - 1; //当获得最优值时，ans记录最后的情况，用于输出路径，flag更新为false，表明目前只有一种方式取得最优值
            flag = false; return true;   //返回true，表明找到最优值
        }
        else if (now == val) flag = true;//flag为true表示当前最优值可通过多种方式获得
        return false;
    }

    bool mark = false;
    for (int i = 1; i <= len - id; i++) {
        int cur = id+i-1;
        if (dfs (id + i, now + num[id][cur], id)) {//返回值为true时(即找到当前最优指时)记录路径
            pre[id][cur].l = last; pre[id][cur].r = id - 1;
            mark = true;
        }
    }
    return mark;
}

void output(int l, int r) //递归输出路径
{
    if (l < 0 && r < 0) return;
    output(pre[l][r].l, pre[l][r].r);
    printf ("%d ", num[l][r]);
}

int main()
{
    char tmp[10];

    while (scanf ("%d %s", &tar, tmp)) {
         if (!tar && tmp[0] == ‘0‘) break;
         memset(num, 0, sizeof (num));
         convert(tmp);
         preprocess();

         flag = false;
         val = -1;
         dfs(0, 0, -1);
         if (flag) {
            printf ("rejected\n");
            continue;
         }
         if (val == -1) printf ("error\n");
         else {
            printf ("%d ", val);
            output (ans.l, ans.r);
            printf ("\n");
         }
    }
    return 0;
}