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;
#pragma comment(linker, "/STACK:1024000000,1024000000")
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
int gcd(int a, int b) {
  while (b) {
    int t = a % b;
    a = b;
    b = t;
  }
  return a;
}
long long a[901000], sum[2010][2010], mi1[2010][2010], id[2010][2010];
int main() {
  int n, x;
  scanf("%d%d", &n, &x);
  memset(mi1, 0x3f, sizeof mi1);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == j)
        mi1[i][j] = a[i], id[i][j] = i;
      else {
        if (mi1[i][j - 1] < a[j]) {
          mi1[i][j] = a[j];
          id[i][j] = j;
        } else
          mi1[i][j] = mi1[i][j - 1], id[i][j] = id[i][j - 1];
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < n; j++) {
      int t = (n + i - 1) / i;
      int k, d;
      for (k = j, d = 0; d < t - 1; d++, k += i, k %= n) {
        sum[j][i - 1] += a[k] * i;
      }
      sum[j][i - 1] += 1LL * (id[(j - i + n) % n][k] - (k - i + n) % n) *
                           mi1[(j - i + n) % n][k] -
                       1LL * (j - id[(j - i + n) % n][k] + n) % n * a[j];
    }
  }
  long long ans = 1e18;
  for (int i = 0; i < n; i++) {
    long long mi = 1e18;
    for (int j = 0; j < n; j++) mi = min(sum[j][i], mi);
    ans = min(ans, mi + 1LL * i * x);
  }
  cout << ans << '\n';
}
