#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, unsigned long long int y,
                    long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int mI(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int n, k;
vector<long long int> v;
long long int dp[300001][3][3];
long long int fun(long long int i, long long int s, long long int s1) {
  if (i >= n) return 0;
  if (dp[i][s][s1] != (long long int)1e18) return dp[i][s][s1];
  long long int ans = -1e18;
  if (s == 0) {
    ans = max(ans, v[i] + fun(i + 1, 1, 0));
    ans = max(ans, fun(i + 1, 0, 0));
    ans = max(ans, k * v[i] + fun(i + 1, 1, 1));
  }
  ans = max(ans, fun(i + 1, 2, 2));
  if (s == 1) {
    if (s1 == 0) {
      ans = max(ans, v[i] + fun(i + 1, 1, 0));
      ans = max(ans, k * v[i] + fun(i + 1, 1, 1));
    } else if (s1 == 1) {
      ans = max(ans, k * v[i] + fun(i + 1, 1, 1));
      ans = max(ans, v[i] + fun(i + 1, 1, 2));
    } else if (s1 == 2) {
      ans = max(ans, v[i] + fun(i + 1, 1, 2));
    }
  }
  return dp[i][s][s1] = ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  v.resize(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (long long int i = 0; i < 300001; i++) {
    for (long long int j = 0; j < 3; j++) {
      for (long long int k = 0; k < 3; k++) dp[i][j][k] = (long long int)1e18;
    }
  }
  cout << fun(0, 0, 0);
}

### Prompt

Develop a solution in Cpp to the problem described below:

You are given an array a consisting of n 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 a and multiply all values contained in this subarray by x. You want to maximize the beauty of array after applying at most one such operation.

Input

The first line contains two integers n and x (1 ≤ n ≤ 3 ⋅ 10^5, -100 ≤ x ≤ 100) — the length of array a and the integer x respectively.

The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the array a.

Output

Print one integer — the maximum possible beauty of array a 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, unsigned long long int y,
                    long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int mI(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int n, k;
vector<long long int> v;
long long int dp[300001][3][3];
long long int fun(long long int i, long long int s, long long int s1) {
  if (i >= n) return 0;
  if (dp[i][s][s1] != (long long int)1e18) return dp[i][s][s1];
  long long int ans = -1e18;
  if (s == 0) {
    ans = max(ans, v[i] + fun(i + 1, 1, 0));
    ans = max(ans, fun(i + 1, 0, 0));
    ans = max(ans, k * v[i] + fun(i + 1, 1, 1));
  }
  ans = max(ans, fun(i + 1, 2, 2));
  if (s == 1) {
    if (s1 == 0) {
      ans = max(ans, v[i] + fun(i + 1, 1, 0));
      ans = max(ans, k * v[i] + fun(i + 1, 1, 1));
    } else if (s1 == 1) {
      ans = max(ans, k * v[i] + fun(i + 1, 1, 1));
      ans = max(ans, v[i] + fun(i + 1, 1, 2));
    } else if (s1 == 2) {
      ans = max(ans, v[i] + fun(i + 1, 1, 2));
    }
  }
  return dp[i][s][s1] = ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  v.resize(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (long long int i = 0; i < 300001; i++) {
    for (long long int j = 0; j < 3; j++) {
      for (long long int k = 0; k < 3; k++) dp[i][j][k] = (long long int)1e18;
    }
  }
  cout << fun(0, 0, 0);
}

```