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>
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int vx[4] = {0, 1, 0, -1}, vy[4] = {1, 0, -1, 0};
using namespace std;
int main() {
  long long N, X;
  cin >> N >> X;
  long long A[2 * N + 1], ANS[N + 1];
  for (int i = 1; i <= N; ++i) {
    cin >> A[i];
    A[N + i] = A[i];
    ANS[i] = i * X;
  }
  long long DP[N + 1][N + 1];
  for (int i = (0); i < (N + 1); ++i) {
    for (int j = (0); j < (N + 1); ++j) {
      DP[i][j] = 0LL;
      if (j == 0 && i > 0) DP[i][j] = A[i];
    }
  }
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N - 1; ++j) {
      DP[i][j] = min(DP[i][j - 1], A[i + N - j]);
    }
  }
  for (int i = 1; i <= N; ++i) {
    for (int j = 0; j <= N - 1; ++j) {
      ANS[j] += DP[i][j];
    }
  }
  cout << *min_element(ANS + 0, ANS + N) << '\n';
  return 0;
}
