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 sz = 4007;
int n;
int cst;
int a[sz];
long long ans;
deque<int> q;
void insert(int x) {
  while (!q.empty() && a[q.back()] > a[x]) q.pop_back();
  q.push_back(x);
}
int main() {
  ans = 1e18;
  scanf("%d%d", &n, &cst);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[i + n] = a[i];
  for (int k = 0; k <= n - 1; ++k) {
    long long sum = 0;
    while (q.size()) q.pop_back();
    for (int i = n - k + 1; i <= n; ++i) insert(i);
    for (int i = n + 1; i <= 2 * n; ++i) {
      while (!q.empty() && i - q.front() > k) q.pop_front();
      insert(i);
      sum += a[q.front()];
    }
    ans = min(ans, sum + 1ll * k * cst);
  }
  printf("%d\n", ans);
}
