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 N = 2000;
const long long INF = (long long)1e18;
int n, ant[N], tol[N];
long long x, a[N], sol;
bitset<N> act;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ant[i] = i - 1;
  }
  ant[0] = n - 1;
  while (act.count() < n) {
    queue<int> q;
    for (int i = 0; i < n; i++) {
      if (act[i] && act[ant[i]] == 0) {
        tol[i] = i;
        q.push(i);
      } else {
        tol[i] = -1;
      }
    }
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      int j = (i + 1) % n;
      if (tol[j] == -1) {
        tol[j] = tol[i];
        q.push(j);
      }
    }
    for (int i = 0; i < n; i++) {
      int j = ant[i];
      if (act[j]) {
        j = tol[j];
        a[i] = min(a[i], x + a[j]);
      }
    }
    long long mn = INF;
    for (int i = 0; i < n; i++) {
      if (act[i] == 0) {
        mn = min(mn, a[i]);
      }
    }
    for (int i = 0; i < n; i++) {
      if (act[i] == 0 && mn == a[i]) {
        sol += mn;
        act[i] = 1;
      }
    }
  }
  cout << sol << "\n";
}
