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;
long long N, X, A[2000];
int main(void) {
  cin >> N >> X;
  for (long long i = ((long long)0); i < ((long long)N); i++) cin >> A[i];
  vector<vector<pair<long long, long long> > > cost(
      N, vector<pair<long long, long long> >(N));
  for (long long i = ((long long)0); i < ((long long)N); i++) {
    cost[i][0] = pair<long long, long long>(A[i], 0);
    for (long long j = ((long long)1); j < ((long long)N); j++) {
      long long cur = (i - j + N) % N;
      cost[i][j] =
          min(cost[i][j - 1], pair<long long, long long>(A[cur] + X * j, j));
    }
  }
  long long ans = (1LL << 60);
  for (long long i = ((long long)0); i < ((long long)N); i++) {
    long long sum = 0;
    for (long long j = ((long long)0); j < ((long long)N); j++)
      sum += cost[j][i].first - X * cost[j][i].second;
    sum += X * i;
    ans = min(ans, sum);
  }
  cout << ans << endl;
}
