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 MAX = 2000 + 10;
const int INF = 1000000000 + 1;
int N, x;
int a[MAX];
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; ++i) cin >> a[i];
  bool dir[MAX] = {0};
  for (int i = 0; i < N; ++i) {
    if (i == 0) {
      if (a[N - 1] < x) {
        if (a[N - 1] + x <= a[0]) dir[i] = 1;
      } else if (x <= a[0])
        dir[i] = 1;
    } else {
      if (a[i - 1] < x) {
        if (a[i - 1] + x <= a[i]) dir[i] = 1;
      } else if (x <= a[i])
        dir[i] = 1;
    }
  }
  int n_max_shift = 0;
  int n_shift = 0;
  int p = 0;
  while (p < N) {
    if (dir[p] == 0) break;
    ++p;
  }
  for (int i = (p + 1) % N; i != p; i = (i + 1) % N) {
    if (dir[i] == 1) {
      ++n_shift;
      n_max_shift = max(n_shift, n_max_shift);
    } else
      n_shift = 0;
  }
  int cost[MAX];
  for (int i = 0; i < N; ++i) {
    cost[i] = a[i];
    int j = i - 1;
    if (j < 0) j = N - 1;
    int t = 0;
    while (t < n_max_shift) {
      cost[i] = min(cost[i], a[j]);
      --j;
      if (j < 0) j = N - 1;
      ++t;
    }
  }
  long long res = n_max_shift * x;
  for (int i = 0; i < N; ++i) res += cost[i];
  cout << res << endl;
  return 0;
}
