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 int maxn = 2e3 + 5;
long long a[maxn];
int vis[maxn];
int main() {
  int n;
  long long x, sum = 0, mi = 1LL << 60LL;
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mi = min(mi, a[i]);
  }
  int all = 0, pos;
  for (int i = 0; i < n; i++)
    if (a[i] == mi) vis[i] = 1, sum += a[i], all++;
  while (true) {
    bool flag = 0;
    if (all == n) break;
    all = 0;
    for (int i = 0; i < n - 1; i++) {
      if (vis[i] == 1 && vis[i + 1] == 0 && a[i + 1] <= x)
        sum += a[i + 1], vis[i + 1] = 1;
    }
    if (vis[n - 1] == 1 && vis[0] == 0 && a[0] <= x) sum += a[0], vis[0] = 1;
    for (int i = 0; i < n; i++)
      if (vis[i] == 1) all++;
    if (all == n) break;
    all = 0;
    sum += x;
    for (int i = n - 1; i >= 0; i--) {
      if (vis[i] == 1) {
        if (i == n - 1)
          flag = 1;
        else
          vis[i] = 0, vis[i + 1] = 1;
      }
    }
    if (flag) vis[0] = 1;
    for (int i = 0; i < n; i++) {
      if (a[i] == mi && vis[i] == 0) sum += a[i], vis[i] = 1;
      if (vis[i] == 1) all++;
    }
  }
  cout << sum << endl;
}
