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;
template <class T, class U>
void cmax(T& a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
void cmin(T& a, U b) {
  if (a > b) a = b;
}
int n, x;
int pre(int i) {
  if (i == 0) return n - 1;
  return i - 1;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> x;
  vector<int> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  const long long INF = 1LL << 50;
  vector<vector<int>> cnt(n, vector<int>(n, 0));
  vector<vector<long long>> cost(n, vector<long long>(n, INF));
  for (long long i = 0; i < n; i++) cost[0][i] = a[i];
  for (long long i = 0; i < n - 1; i++) {
    for (long long j = 0; j < n; j++) {
      if (cost[i][pre(j)] < cost[i + 1][j] && cost[i][pre(j)] < cost[i][j]) {
        cost[i + 1][j] = cost[i][pre(j)] + x;
        cnt[i + 1][j] = cnt[i][pre(j)] + 1;
      } else if (cost[i][j] < cost[i + 1][j]) {
        cost[i + 1][j] = cost[i][j];
      }
    }
  }
  long long ans = INF;
  for (long long i = 0; i < n; i++) {
    int cntsum = 0;
    int cntmax = 0;
    long long costsum = 0;
    for (long long j = 0; j < n; j++) {
      costsum += cost[i][j];
      cntsum += cnt[i][j];
      cmax(cntmax, cnt[i][j]);
    }
    cmin(ans, costsum - (cntsum - cntmax) * x);
  }
  cout << ans << endl;
}
