/*
* 分组背包问题
* 状态:dp[i][j], 表示前i个科目,使用j天所获得的max gain
* 状态转移方程: dp[i][j] = max(dp[i-1][j], dp[i-1][j-k] + val[i][k])
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 101
using namespace std;
int dp[MAXN][MAXN], val[MAXN][MAXN];
int main(int argc, char const *argv[])
{
//freopen("test.in", "r", stdin);
int cls_cnt, day_cnt, ans;
while( scanf("%d %d", &cls_cnt, &day_cnt) ) {
if( !cls_cnt && !day_cnt ) {
break;
}
for(int i = 1; i <= cls_cnt; i ++) {
for(int j = 1; j <= day_cnt; j ++) {
scanf("%d", &val[i][j]);
}
}
ans = 0;
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= cls_cnt; i ++) {
for(int j = day_cnt; j >= 1; j --) {
for(int k = 1; k <= j; k ++) {
dp[i][j] = max(dp[i-1][j-k]+val[i][k], dp[i][j]);
}
dp[i][j] = max(dp[i][j], dp[i-1][j]);
ans = max(ans, dp[i][j]);
}
}
printf("%d\n", ans);
}
return 0;
}