Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

* Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

* Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.




Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

* 2≤N≤2,000
* a_i are integers.
* 1≤a_i≤10^9
* x is an integer.
* 1≤x≤10^9

Input

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


N x
a_1 a_2 ... a_N


Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Examples

Input

2 10
1 100


Output

12


Input

3 10
100 1 100


Output

23


Input

4 10
1 2 3 4


Output

10
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-10;
const int INF = 2147483647;
int N;
int m, x;
int a[2000];
long long s[2000][2000][2];
long long dp[2001][2];
int main() {
  scanf("%d%d", &N, &x);
  m = 0;
  for (int i = 0; i < N; ++i) {
    scanf("%d", a + i);
    if (a[m] > a[i]) m = i;
  }
  for (int i = 0; i < N; ++i) {
    s[i][0][0] = a[i];
    s[i][0][1] = 0;
    for (int j = 1; j < N; ++j) {
      if (a[(i + j) % N] > x + a[i]) {
        s[i][j][0] = s[i][j - 1][0] + a[i];
        s[i][j][1] = s[i][j - 1][1] + 1;
      } else {
        s[i][j][0] = s[i][j - 1][0] + a[(i + j) % N];
        s[i][j][1] = s[i][j - 1][1];
      }
    }
  }
  dp[0][0] = a[m];
  dp[0][1] = 0;
  for (int i = 1; i < N; ++i) {
    dp[i][0] = s[m][i][0];
    dp[i][1] = s[m][i][1];
    for (int j = 0; j < i; ++j) {
      int y = ((dp[j][1]) > (s[(m + j + 1) % N][i - j - 1][1])
                   ? (dp[j][1])
                   : (s[(m + j + 1) % N][i - j - 1][1]));
      if (dp[i][0] + dp[i][1] * x >
          dp[j][0] + s[(m + j + 1) % N][i - j - 1][0] + x * y) {
        dp[i][0] = dp[j][0] + s[(m + j + 1) % N][i - j - 1][0];
        dp[i][1] = y;
      }
    }
  }
  printf("%lld\n", dp[N - 1][0] + dp[N - 1][1] * x);
  return 0;
}
