Educational Codeforces Round 63 D. Beautiful Array

D. Beautiful Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array aa consisting of nn integers. Beauty of array is the maximum sum of some consecutive subarray of this array (this subarray may be empty). For example, the beauty of the array [10, -5, 10, -4, 1] is 15, and the beauty of the array [-3, -5, -1] is 0.

You may choose at most one consecutive subarray of aa and multiply all values contained in this subarray by xx. You want to maximize the beauty of array after applying at most one such operation.

Input

The first line contains two integers nn and xx (1≤n≤3⋅105,−100≤x≤100) — the length of array aa and the integer xx respectively.

The second line contains nn integers a1,a2,…,an (−109≤ai≤109) — the array aa.

Output

Print one integer — the maximum possible beauty of array aa after multiplying all values belonging to some consecutive subarray x.

Examples

input

5 -2
-3 8 -2 1 -6

output

22

input

12 -3
1 3 3 7 1 3 3 7 1 3 3 7

output

42

input

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

output

0

Note

In the first test case we need to multiply the subarray [-2, 1, -6], and the array becomes [-3, 8, 4, -2, 12] with beauty 22([-3, 8, 4, -2, 12]).

In the second test case we don't need to multiply any subarray at all.

In the third test case no matter which subarray we multiply, the beauty of array will be equal to 0.

 果然还是太菜了，简单dp都想不出来。

dp[0] 记录不使用x的最大值

dp[1] 记录允许使用x的情况下的最大值

dp[2]记录在前面的区间中已经使用了x的情况下的最大值(尽管可能使用x的区间对最大值并没有贡献)

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

long long arr[300005];

int main(){
    int n, x;
    long long dp[3] = {},ans=0;
    scanf("%d%d",&n,&x);
    for(int i=0;i<n;i++){
        scanf("%I64d",&arr[i]);
        dp[0] = max(0ll,dp[0]+arr[i]);
        dp[1] = max(dp[0],dp[1]+arr[i]*x);
        dp[2] = max(dp[1],dp[2]+arr[i]);
        ans = max(ans,dp[2]);

    }
    printf("%I64d
",ans);
    return 0;

}