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;
int N, X;
unsigned long long a[2000][2000];
int main(int argc, char **argv) {
  unsigned long long min[2000];
  unsigned long long cost[2000];
  cin >> N;
  cin >> X;
  for (int i = 0; i < N; i++) {
    cin >> a[0][i];
    min[i] = a[0][i];
    cost[i] = a[0][i];
  }
  int index = 0;
  for (int i = 1; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (j) {
        if (a[i - 1][j] < UINT64_MAX - X)
          a[i][j] = a[i - 1][j - 1] + X;
        else
          a[i][j] = UINT64_MAX;
      } else {
        if (a[i - 1][j] < UINT64_MAX - X)
          a[i][j] = a[i - 1][N - 1] + X;
        else
          a[i][j] = UINT64_MAX;
      }
      if (a[i][j] < min[j]) {
        min[j] = a[i][j];
        if (j >= i)
          cost[j] = a[0][j - i];
        else
          cost[j] = a[0][j + N - i];
        if (index < i) index = i;
      }
    }
  }
  unsigned long long ans = 0;
  for (int i = 0; i < N; i++) {
    ans += cost[i];
  }
  cout << ans + X * index;
  return 0;
}
