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>
inline void inp(long long int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  n = n * sign;
}
using namespace std;
long long int n, x;
const int MAXN = (long long int)2e3 + 10;
long long int arr[MAXN];
long long int dp[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> x;
  long long int poesreq = -1;
  long long int mintillnow = (long long int)1e18;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] <= mintillnow) {
      mintillnow = arr[i];
      poesreq = i;
    }
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    if (arr[i] > x && mintillnow < x) {
      ans += (x + abs(poesreq - i));
    } else {
      ans += arr[i];
    }
  }
  cout << ans;
  return 0;
}
