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 = 2010;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
int a[maxn];
int n;
long long x, d[2 * maxn][maxn];
deque<int> q;
int main() {
  scanf("%d %lld", &n, &x);
  for (int i = (1); i <= (n); ++i) scanf("%d", a + i), a[n + i] = a[i];
  memset(d, INF, sizeof d);
  ;
  long long ans = INF;
  for (int k = (0); k < (n); ++k) {
    while (q.size()) q.pop_back();
    long long res = x * k;
    int r = k + 1;
    for (int i = (1); i <= (r); ++i) {
      while (!q.empty() && a[q.back()] > a[i]) q.pop_back();
      q.push_back(i);
    }
    while (r <= n + k) {
      res += a[q.front()];
      if (a[q.front()] == a[r - k]) q.pop_front();
      ++r;
      while (!q.empty() && a[q.back()] > a[r]) q.pop_back();
      q.push_back(r);
    }
    ans = min(ans, res);
  }
  printf("%lld\n", ans);
  return 0;
}
