Problem:
Score : 500 points

Problem Statement

There are K items placed on a grid of squares with R rows and C columns. Let (i, j) denote the square at the i-th row (1 \leq i \leq R) and the j-th column (1 \leq j \leq C). The i-th item is at (r_i, c_i) and has the value v_i.

Takahashi will begin at (1, 1), the start, and get to (R, C), the goal. When he is at (i, j), he can move to (i + 1, j) or (i, j + 1) (but cannot move to a non-existent square).

He can pick up items on the squares he visits, including the start and the goal, but at most three for each row. It is allowed to ignore the item on a square he visits.

Find the maximum possible sum of the values of items he picks up.

Constraints

1 \leq R, C \leq 3000

1 \leq K \leq \min(2 \times 10^5, R \times C)

1 \leq r_i \leq R

1 \leq c_i \leq C

(r_i, c_i) \neq (r_j, c_j) (i \neq j)

1 \leq v_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

R C K
r_1 c_1 v_1
r_2 c_2 v_2
:
r_K c_K v_K

Output

Print the maximum possible sum of the values of items Takahashi picks up.

Sample Input 1

2 2 3
1 1 3
2 1 4
1 2 5

Sample Output 1

8

He has two ways to get to the goal:

Visit (1, 1), (1, 2), and (2, 2), in this order. In this case, the total value of the items he can pick up is 3 + 5 = 8.

Visit (1, 1), (2, 1), and (2, 2), in this order. In this case, the total value of the items he can pick up is 3 + 4 = 7.

Thus, the maximum possible sum of the values of items he picks up is 8.

Sample Input 2

2 5 5
1 1 3
2 4 20
1 2 1
1 3 4
1 4 2

Sample Output 2

29

We have four items in the 1-st row. The optimal choices are as follows:

Visit (1, 1) (1, 2), (1, 3), (1, 4), (2, 4), and (2, 5), in this order, and pick up all items except the one on (1, 2). Then, the total value of the items he picks up will be 3 + 4 + 2 + 20 = 29.

Sample Input 3

4 5 10
2 5 12
1 5 12
2 3 15
1 2 20
1 1 28
2 4 26
3 2 27
4 5 21
3 5 10
1 3 10

Sample Output 3

142

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define vi vector<int>
#define vll vector<ll>
#define vvi vector < vi >
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vpll vector < pair<long long,long long> >
#define fill(a,val)  memset(a,val, sizeof (a))
#define sort_unique(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define all(cont) cont.begin(), cont.end()
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define max4(a,b,c,d) max(max(a,d),max(b,c))
#define MAX   1e9
#define MIN  -1e9
typedef map<int,int> mpi;
typedef set<int> seti;
typedef multiset<int> mseti;
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define rep(i, n)    for(long long int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(long long int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(long long int i = a; i >= (n); --i)
#define repn(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define trav(a, x) for(auto& a : x)
#define dbg(x) cout << #x << " = " << x << '\n'
#define dbg2(x,y) cout<<#x<<" = "<<x<<" & "<<#y<<" = "<<y<<endl
#define inp(a,start,len) for(long long int i = start;i < len;i++) cin>>a[i]
struct Interval{ll start,end;};
bool compare(Interval i1, Interval i2) {return (i1.start < i2.start);}

ll dp[3002][3002][4];

int main(){
	fio;
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	ll R,C,K;
	cin>>R>>C>>K;
	ll v[R+1][C+1];
	fill(v,0);
	// dbg(v[1][1]);
	rep(i,K){
		ll r,c,val;
		cin>>r>>c>>val;
		v[r][c] = val;
	}
	repA(i,1,R){
		repA(j,1,C){
			ll up = max4(dp[i-1][j][0],dp[i-1][j][1],dp[i-1][j][2],dp[i-1][j][3]);
			if(!v[i][j]){
				rep(k,4){
					dp[i][j][k] = max(dp[i][j-1][k],up);
				}
			}
			else{
				dp[i][j][0] = max(dp[i][j-1][0],up);
				dp[i][j][1] = max3(dp[i][j-1][0]+v[i][j],up+v[i][j],dp[i][j-1][1]);
				dp[i][j][2] = max3(dp[i][j-1][1]+v[i][j],dp[i][j][1],dp[i][j-1][2]);
				dp[i][j][3] = max3(dp[i][j-1][2]+v[i][j],dp[i][j][2],dp[i][j-1][3]);
			}
		}
	}
	cout<<max4(dp[R][C][0],dp[R][C][1],dp[R][C][2],dp[R][C][3]);
	return 0;
}