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 <typename P, typename Q>
ostream& operator<<(ostream& os, pair<P, Q> p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
const int N = 2000 + 10;
long long int cost[N][N];
int main(int argc, char* argv[]) {
  int n, x;
  while (cin >> n >> x) {
    long long int a[n];
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    fill(&cost[0][0], &cost[N - 1][N - 1] + 1, 0);
    const long long int inf = 1LL << 60;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        cost[i][j] += a[j];
      }
    }
    for (int i = 0; i < n; ++i) {
    }
    long long int dp[n];
    for (int i = 0; i < n; ++i) {
      dp[i] = a[i];
    }
    long long int mn = inf;
    for (int ench = 0; ench <= 2 * n; ++ench) {
      for (int slime = 0; slime < n; ++slime) {
        const int i = (slime + ench) % n;
        dp[slime] = min(dp[slime], cost[slime][i]);
      }
      mn = min(mn, accumulate(dp, dp + n, 0LL) + ench * x);
    }
    cout << mn << endl;
  }
  return 0;
}
