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>
int main() {
  long long n, x;
  std::cin >> n >> x;
  std::vector<long long> dp(n);
  std::vector<int> w(n, 0);
  std::queue<int> update;
  for (int i = 0; i < n; i++) {
    std::cin >> dp[i];
    update.push(i);
    w[i] = i;
  }
  while (!update.empty()) {
    int pos = update.front();
    update.pop();
    int next = (pos + 1) % n;
    if (dp[next] > dp[pos] + x) {
      dp[next] = dp[pos] + x;
      update.push(next);
      w[next] = w[pos];
    }
  }
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    sum += dp[w[i]];
  }
  int start = 0;
  for (int i = 0; i < n; i++) {
    if (w[i] != w[(i + 1) % n]) start = (i + 1) % n;
  }
  long long w_max = 0, count = 0, num = 0, p = start;
  while (true) {
    int pre = p;
    while (true) {
      int next = (p + 1) % n;
      count++;
      if (w[p] != w[next]) break;
      p = next;
      if (p == start) {
        p--;
        break;
      }
    }
    if (pre == p) p++;
    w_max = std::max(w_max, count);
    count = 0;
    if ((p + 1) % n == start) break;
    p %= n;
  }
  std::cout << sum + x * (w_max - 1) << std::endl;
  return 0;
}
