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 mypair {
  long long int a;
  long long int count;
  mypair() {}
  mypair(int i, int j) : a(i), count(j) {}
};
long long int const MOD = 1000000007;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long int k = 0;
  long long int n, x;
  cin >> n >> x;
  vector<vector<mypair>> a(n, vector<mypair>(n));
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[0][i] = mypair(x, 0);
  }
  long long int min = 0;
  for (int i = 0; i < n; i++) {
    min += a[0][i].a;
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (j == 0) {
        if (a[i - 1][j].a > a[i - 1][n - 1].a + x) {
          a[i][j].a = a[i - 1][n - 1].a + x;
          a[i][j].count = a[i - 1][n - 1].count + 1;
        } else {
          a[i][j] = a[i - 1][j];
        }
      } else {
        if (a[i - 1][j].a > a[i - 1][j - 1].a + x) {
          a[i][j].a = a[i - 1][j - 1].a + x;
          a[i][j].count = a[i - 1][j - 1].count + 1;
        } else {
          a[i][j] = a[i - 1][j];
        }
      }
    }
    long long int mcount = 0;
    long long int tmpsum = 0;
    for (int j = 0; j < n; j++) {
      if (a[i][j].count > mcount) {
        mcount = a[i][j].count;
      }
      tmpsum += a[i][j].a - (a[i][j].count * x);
    }
    tmpsum += mcount * x;
    if (tmpsum < min) {
      min = tmpsum;
    }
  }
  cout << min << endl;
  return 0;
}
