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;
struct node {
  long long mllime;
  long long count = 0;
  long long a;
} M[2010];
int main(void) {
  long long N, x, a[2010];
  cin >> N >> x;
  for (long long i = 0; i < N; i++) {
    scanf("%lld", &a[i]);
    M[i].mllime = a[i];
    M[i].a = a[i];
  }
  for (long long i = 0; i < N - 1; i++) {
    long long sum = (N - 1 - i) * x + a[i];
    if (sum < M[N - 1].mllime) {
      M[N - 1].mllime = sum;
      M[N - 1].count = N - 1 - i;
      M[N - 1].a = a[i];
    }
  }
  for (long long i = 0; i < N - 1; i++) {
    if (i == 0) {
      if (M[N - 1].mllime + x < M[0].mllime) {
        M[0].mllime = M[N - 1].mllime + x;
        M[i].count = M[N - 1].count + 1;
        M[i].a = M[N - 1].a;
      }
    } else {
      if (M[i - 1].mllime + x < M[i].mllime) {
        M[i].mllime = M[i - 1].mllime + x;
        M[i].count = M[i - 1].count + 1;
        M[i].a = M[i - 1].a;
      }
    }
  }
  long long ans = 0, maxCount = 0;
  for (long long i = 0; i < N; i++) {
    if (M[i].count > maxCount) maxCount = M[i].count;
    ans += M[i].a;
  }
  ans += maxCount * x;
  cout << ans << endl;
}
