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 MX = 2e2 + 9;
const long double eps = -0.00000001;
const long long mod = 1e9 + 7;
long long po(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long x = po(a, b / 2) % mod;
  x *= x;
  if (b % 2) {
    return x * a;
  }
  return x;
}
int get(string x) {
  int c = 0;
  for (int i = 0; i < x.size(); i++) {
    c = c * 10 + x[i] - '0';
  }
  return c;
}
int dx[] = {0, 0, 1, -1};
int dy[] = {-1, 1, 0, 0};
int n, arr[MX], x;
long long dp[2009][2009], ans;
int main() {
  scanf("%d %d", &n, &x);
  for (int j = 0; j < n; j++) {
    scanf("%d", &arr[j]);
    dp[0][j] = arr[j];
    ans += arr[j];
  }
  for (int j = 1; j < n; j++) {
    long long cur = 0;
    for (int i = 0; i < n; i++) {
      dp[j][i] = min(dp[j - 1][i], 1ll * arr[(i - j + n) % n]);
      cur += dp[j][i];
    }
    cur += 1ll * x * j;
    ans = min(ans, cur);
  }
  cout << ans << endl;
}
