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 = 5000;
int n;
long long p[MAXN], dp[MAXN][MAXN], x;
int mini[MAXN][MAXN];
long long rek(int l, int r) {
  if (r < l) return 0;
  if (l == r) return p[l];
  if (dp[l][r] != -1) return dp[l][r];
  long long ret = (r - l + 1) * p[l] + (r - l) * x;
  ret = min(ret, p[l] + rek(l + 1, r));
  if (mini[l][r] != l) {
    long long L = rek(l, mini[l][r] - 1);
    long long R = rek(mini[l][r], r);
    ret = min(ret, L + R);
  }
  return dp[l][r] = ret;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d %lld", &n, &x);
  for (int i = (0); i < int(n); i++) scanf("%lld", &p[i]);
  for (int i = (0); i < int(n); i++) p[i + n] = p[i];
  for (int i = (0); i < int(2 * n); i++) {
    int tmp = i;
    for (int j = (i); j < int(2 * n); j++) {
      if (p[j] < p[tmp]) tmp = j;
      mini[i][j] = tmp;
    }
  }
  long long sol = (1LL << 60);
  for (int i = (0); i < int(2 * n); i++) {
    for (int j = (0); j < int(i + 1); j++) {
      if (i - j + 1 == n) sol = min(sol, rek(j, i));
    }
  }
  printf("%lld\n", sol);
  return 0;
}
