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# include <bits/stdc++.h>
using namespace std ;
typedef long long LL ;
const int N = 1010 ;
int g [ N ] [ N ] ; // 原始地图
// 1e3 * 1e3 * 1e4 开longlong
LL f1 [ N ] [ N ] ; // 向右+向下走
LL f2 [ N ] [ N ] ; // 向右+向上走
int n , m ;
/*
现象：
实测 max( max(a,b),max(c,d)) 795 ms
max({a,b,c,d}) 1200ms
结论：
放弃max({a,b,c,d}),这个太慢了
*/
int main ( ) {
// 加快读入
ios : : sync_with_stdio ( false ) , cin . tie ( 0 ) ;
cin > > n > > m ;
for ( int i = 1 ; i < = n ; i + + )
for ( int j = 1 ; j < = m ; j + + )
cin > > g [ i ] [ j ] ;
memset ( f1 , - 0x3f , sizeof f1 ) ; // 有负数，初始化为-INF
memset ( f2 , - 0x3f , sizeof f2 ) ; // 有负数，初始化为-INF
f1 [ 1 ] [ 1 ] = f2 [ 1 ] [ 1 ] = g [ 1 ] [ 1 ] ; // 左上角的数必须拿起来
for ( int j = 1 ; j < = m ; j + + ) { // 第一层循环列数
// 用上一列两种问题的状态更新这一列的状态达到最优 //行数之间也要更新最优的状态
for ( int i = 1 ; i < = n ; i + + ) // 因为是向下走，所以从第一行开始循环
f1 [ i ] [ j ] = max ( max ( f1 [ i ] [ j - 1 ] , f2 [ i ] [ j - 1 ] ) + g [ i ] [ j ] , max ( f1 [ i ] [ j ] , f1 [ i - 1 ] [ j ] + g [ i ] [ j ] ) ) ;
for ( int i = n ; i > = 1 ; i - - ) // 因为是向上走，
f2 [ i ] [ j ] = max ( max ( f1 [ i ] [ j - 1 ] , f2 [ i ] [ j - 1 ] ) + g [ i ] [ j ] , max ( f2 [ i ] [ j ] , f2 [ i + 1 ] [ j ] + g [ i ] [ j ] ) ) ;
}
cout < < f1 [ n ] [ m ] < < endl ;
// cout << max(f1[n][m], f2[n][m]);
// 两种问题取最优（ ， ，
return 0 ;
}
