[CF1433F] Zero Remainder Sum

原题

You are given a matrix aa of size n×mn×m consisting of integers.

You can choose no more than \(⌊\frac{m}{2}⌋\) elements in each row. Your task is to choose these elements in such a way that their sum is divisible by \(k\) and this sum is the maximum.

In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by \(k\).

Note that you can choose zero elements (and the sum of such set is \(0\)).

Input

The first line of the input contains three integers nn, mm and $k (1≤n,m,k≤70) $— the number of rows in the matrix, the number of columns in the matrix and the value of \(k\). The next nn lines contain mm elements each, where the \(j\)-th element of the \(i\)-th row is \(a_{i,j }(1≤a_{i,j}≤70)\).

Output

Print one integer — the maximum sum divisible by \(k\) you can obtain.

Examples

input

3 4 3
1 2 3 4
5 2 2 2
7 1 1 4

output

24

input

5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6

output

56

Note

In the first example, the optimal answer is \(2\) and \(4\) in the first row, \(5\) and \(2\) in the second row and \(7\) and $4 $ in the third row. The total sum is \(2+4+5+2+7+4=24\).

思路

开四维数组 dp[行][列][当前行选了几个][余数]，把每行每个余数的dp最大值用mx[]存下来作为下一行选取0个的dp值，最后输出余数为0的最大值。

注意不能取的情况continue掉。

选取的个数从0开始，否则会遗漏掉枚举的这个数作为选取的第一个数的情况，这里卡了好久。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <list>
#include <map>
#include <iostream>
#include <iomanip>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
#define LL long long
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f
#define PI 3.1415926535898
#define F first
#define S second
#define endl '\n'
#define lson  rt << 1
#define rson  rt << 1 | 1
#define lowbit(x) (x &(-x))
#define f(x, y, z) for (int x = (y), __ = (z); x < __; ++x)
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;

const int maxn = 77;
int n, m, k;
int a[maxn][maxn], mx[maxn], dp[maxn][maxn][maxn][maxn];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m >> k;
	_rep(i, 1, n) {
		_rep(j, 1, m) cin >> a[i][j];
	}
	_rep(i, 1, n) {
		_rep(j, 0, m) {
			if (!j) {
				_rep(r, 0, k - 1) dp[i][j][0][r] = mx[r];
				continue;
			}
			_rep(p, 0, min(j, m / 2)) {
				_rep(r, 0, k - 1) {
					dp[i][j][p][(r + a[i][j]) % k] = max(dp[i][j][p][(r + a[i][j]) % k], dp[i][j - 1][p][(r + a[i][j]) % k]);
					if (!(p == 1 && !r) && !dp[i][j - 1][p - 1][r]) continue;
					dp[i][j][p][(r + a[i][j]) % k] = max(dp[i][j][p][(r + a[i][j]) % k], dp[i][j - 1][p - 1][r] + a[i][j]);
					mx[(r + a[i][j]) % k] = max(mx[(r + a[i][j]) % k], dp[i][j][p][(r + a[i][j]) % k]);
				}
			}
		}
	}
	cout << mx[0] << endl;
}