tj

#include <bits/stdc++.h>

using namespace std;
//完全背包问题
//m作为背包总容量，今天的价格当做背包容量，明日价格当做物品的价值
//dp[i] = j用i元钱去购买商品所能盈利的（去除成本）最大值
const long long N = 1e2 + 10;
const long long M = 1e4 + 10;
long long dp[M];
long long t , n , m , a[N][N];

int main( ) {
	cin >> t >> n >> m;
	
	for (int i = 1 ; i <= t ; i++) {
		for (int j = 1 ; j <= n ; j++) {
			cin >> a[i][j];
		}
	}
	
	for (int k = 1 ; k < t ; k++) {
		memset(dp , 0 , sizeof dp);
		
		for (int i = 1 ; i <= n ; i++) {
			for (int j = a[k][i] ; j <= m ; j++) {
				dp[j] = max(dp[j] , dp[j - a[k][i]] + a[k + 1][i] - a[k][i]);
			}
		}
		m += dp[m];
	}
	cout << m << endl;
	
	return 0;
}