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;
struct ddd {
  int cnt;
  long long num;
} dd;
int main() {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = (0); i < (int)(n); i++) cin >> a[i];
  vector<ddd> dp(n);
  int m = min_element(a.begin(), a.end()) - a.begin();
  dp[m] = ddd{0, a[m]};
  int i = m + 1;
  while (i != m) {
    int bef = i - 1;
    if (i == 0) bef = n - 1;
    if (i == n) {
      i = 0;
      continue;
    }
    if (a[i] >= dp[bef].num + x) {
      dp[i].num = dp[bef].num + x;
      dp[i].cnt = dp[bef].cnt + 1;
    } else {
      dp[i].num = a[i];
      dp[i].cnt = 0;
    }
    i++;
  }
  int M = 0;
  for (int i = (0); i < (int)(n); i++) M = max(M, dp[i].cnt);
  long long ans = 0;
  while (M >= 0) {
    for (int i = (0); i < (int)(n); i++) {
      if (dp[i].cnt == M) {
        ans += dp[i].num - dp[i].cnt * x;
      }
    }
    if (M > 0) ans += x;
    M--;
  }
  cout << ans << endl;
  return 0;
}
