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 + 2;
int n, x, cost[MAXN], rmq[MAXN][20];
long long ans, dp[MAXN][MAXN];
void prep() {
  for (int i = 0; i < n; i++) rmq[i][0] = cost[i];
  for (int i = 1; (1 << i) <= n; i++)
    for (int j = 0; j + (1 << (i - 1)) < n; j++)
      rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << (i - 1))][i - 1]);
}
int get_min(int l, int r) {
  if (l > r) swap(l, r);
  int k = log2(r - l + 1);
  return min(rmq[l][k], rmq[r - (1 << k) + 1][k]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> cost[i];
  prep();
  for (int r = 0; r < n; r++) {
    for (int i = 0; i < n; i++) {
      int mn1, mn2;
      mn1 = mn2 = INT_MAX;
      if (i - r < 0) {
        mn1 = get_min(0, i);
        mn2 = get_min((i - r + n) % n, n - 1);
      } else {
        mn1 = get_min(i - r, i);
      }
      if (i == 0)
        dp[i][r] += min(mn1, mn2);
      else
        dp[i][r] += dp[i - 1][r] + min(mn1, mn2);
    }
  }
  ans = INT_MAX;
  for (int i = 0; i < n; i++) {
    long long nval = dp[n - 1][i] + 1LL * i * x;
    ans = min(ans, nval);
  }
  cout << ans << '\n';
  return 0;
}
