tj

#include <bits/stdc++.h>
#define int long long
using namespace std;
/*
三方向路径d解法
*/
const long long N = 1e3 + 10;
long long n , m , a[N][N];
long long up[N];//up[i][j] 从下往上走到i,j途经数之和的max
long long down[N];
long long dp[N];

signed main( ) {
	cin >> n >> m;
	
	for (int i = 1 ; i <= n ; i++) {
		for (int j = 1 ; j <= m ; j++) {
			scanf("%lld" , & a[i][j]);
		}
	}
	dp[1] = a[1][1];
	
	for (int i = 2 ; i <= n ; i++) dp[i] = dp[i - 1] + a[i][1];
	
	for (int j = 2 ; j < m ; j++) {
		memset(up , 0 , sizeof up);
		memset(down , 0 , sizeof down);
		down[1] = dp[1] + a[1][j];
		up[n] = dp[n] + a[n][j];
		for (int i = 2 ; i <= n ; i++) down[i] = max(down[i - 1] , dp[i]) + a[i][j]; 
		for (int i = n - 1 ; i >= 1 ; i--) up[i] = max(up[i + 1] , dp[i]) + a[i][j];
		for (int i = 1 ; i <= n ; i++) dp[i] = max(up[i] , down[i]);
	}
	dp[1] += a[1][m];
	for (int i = 2 ; i <= n ; i++) {
		dp[i] = max(dp[i] , dp[i - 1]) + a[i][m];
	}
	cout << dp[n] << endl;
	return 0;
}