#include<iostream>
#include<cstring>

using namespace std;
const int N = 105;
int f[N][N];
int w[N];
int n, k;

int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)scanf("%d", &w[i]);

    for (int i = 0; i < k; i++)f[0][i] = -2e7;
    f[0][0] = 0;//只有f[0][i]中只有0有意义，其他没意义，要设为负无穷，否则会报错

    for (int i = 1; i <= n; i++)//i表示从前i个物品当中选
        for (int j = 0; j < k; j++)//j表示这些物品的价值模k等于j因此j是从0到k
            f[i][j] = max(f[i - 1][j], f[i - 1][(j - w[i] % k + k) % k] + w[i]);//求j-w[i]模以k的余数要让w[i]先模k

    printf("%d\n", f[n][0]);//输出从前n个产品中选，且总价是k的余数
    return 0;
}