#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXM = 200;
const int MAXN = 5000;

typedef long long ll;

int N, M;
int lg[MAXN + 5], st[13][MAXM + 5][MAXN + 5];
void get_st() {
	for(int i=2;i<=N;i++)
		lg[i] = lg[i>>1] + 1;
	for(int p=1;p<=M;p++)
		for(int i=1;i<13;i++) {
			int k = (1 << (i - 1));
			for(int j=1;j<=N;j++)
				st[i][p][j] = max(st[i-1][p][j], st[i-1][p][j+k]);
		}
}
ll rmq(int id, int l, int r) {
	int k = lg[r - l + 1], p = (1<<k);
	return max(st[k][id][l], st[k][id][r-p+1]);
}
ll ans, A[MAXN + 5];
void solve(int opl, int opr, int l, int r) {
	if( l > r ) return ;
	int m = (l + r) >> 1, opm = opl;
	ll tmp = 0;
	for(int i=opl;i<=opr&&i<=m;i++) {
		ll x = 0;
		for(int j=1;j<=M;j++)
			x = (x + rmq(j, i, m));
		x -= (A[m] - A[i]);
		if( x > tmp )
			tmp = x, opm = i;
	}
	ans = max(ans, tmp);
	solve(opl, opm, l, m - 1);
	solve(opm, opr, m + 1, r);
}
int main() {
	scanf("%d%d", &N, &M);
	for(int i=2;i<=N;i++) scanf("%lld", &A[i]);
	for(int i=2;i<=N;i++) A[i] += A[i-1];
	for(int i=1;i<=N;i++)
		for(int j=1;j<=M;j++)
			scanf("%d", &st[0][j][i]);
	get_st(), solve(1, N, 1, N);
	printf("%lld\n", ans);
}
### Prompt

In cpp, your task is to solve the following problem:

There are N barbecue restaurants along a street. The restaurants are numbered 1 through N from west to east, and the distance between restaurant i and restaurant i+1 is A_i.

Joisino has M tickets, numbered 1 through M. Every barbecue restaurant offers barbecue meals in exchange for these tickets. Restaurant i offers a meal of deliciousness B_{i,j} in exchange for ticket j. Each ticket can only be used once, but any number of tickets can be used at a restaurant.

Joisino wants to have M barbecue meals by starting from a restaurant of her choice, then repeatedly traveling to another barbecue restaurant and using unused tickets at the restaurant at her current location. Her eventual happiness is calculated by the following formula: "(The total deliciousness of the meals eaten) - (The total distance traveled)". Find her maximum possible eventual happiness.

Constraints

* All input values are integers.
* 2≤N≤5×10^3
* 1≤M≤200
* 1≤A_i≤10^9
* 1≤B_{i,j}≤10^9

Input

The input is given from Standard Input in the following format:


N M
A_1 A_2 ... A_{N-1}
B_{1,1} B_{1,2} ... B_{1,M}
B_{2,1} B_{2,2} ... B_{2,M}
:
B_{N,1} B_{N,2} ... B_{N,M}


Output

Print Joisino's maximum possible eventual happiness.

Examples

Input

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


Output

11


Input

5 3
1 2 3 4
10 1 1
1 1 1
1 10 1
1 1 1
1 1 10


Output

20

### Response

```cpp
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXM = 200;
const int MAXN = 5000;

typedef long long ll;

int N, M;
int lg[MAXN + 5], st[13][MAXM + 5][MAXN + 5];
void get_st() {
	for(int i=2;i<=N;i++)
		lg[i] = lg[i>>1] + 1;
	for(int p=1;p<=M;p++)
		for(int i=1;i<13;i++) {
			int k = (1 << (i - 1));
			for(int j=1;j<=N;j++)
				st[i][p][j] = max(st[i-1][p][j], st[i-1][p][j+k]);
		}
}
ll rmq(int id, int l, int r) {
	int k = lg[r - l + 1], p = (1<<k);
	return max(st[k][id][l], st[k][id][r-p+1]);
}
ll ans, A[MAXN + 5];
void solve(int opl, int opr, int l, int r) {
	if( l > r ) return ;
	int m = (l + r) >> 1, opm = opl;
	ll tmp = 0;
	for(int i=opl;i<=opr&&i<=m;i++) {
		ll x = 0;
		for(int j=1;j<=M;j++)
			x = (x + rmq(j, i, m));
		x -= (A[m] - A[i]);
		if( x > tmp )
			tmp = x, opm = i;
	}
	ans = max(ans, tmp);
	solve(opl, opm, l, m - 1);
	solve(opm, opr, m + 1, r);
}
int main() {
	scanf("%d%d", &N, &M);
	for(int i=2;i<=N;i++) scanf("%lld", &A[i]);
	for(int i=2;i<=N;i++) A[i] += A[i-1];
	for(int i=1;i<=N;i++)
		for(int j=1;j<=M;j++)
			scanf("%d", &st[0][j][i]);
	get_st(), solve(1, N, 1, N);
	printf("%lld\n", ans);
}
```