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>
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 long long int mod = 1e9 + 7, INF = 1e18;
long long int dx[8] = {0, -1, 1, 0, 1, -1, 1, -1},
              dy[8] = {1, 0, 0, -1, -1, -1, 1, 1};
long long int N, X;
long long int arr[2000];
long long int dp[2001][2000];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> N >> X;
  long long int st = INF, st_idx = 0;
  for (long long int i = (0), i_end_ = (N); i < i_end_; i++) {
    cin >> arr[i];
    if (arr[i] < st) {
      st = arr[i];
      st_idx = i;
    }
  }
  reverse(arr, arr + N);
  long long int minv = INF;
  for (long long int i = (0), i_end_ = (N + 1); i < i_end_; i++) {
    for (long long int j = (0), j_end_ = (N); j < j_end_; j++) {
      dp[i][j] = INF;
    }
  }
  dp[1][0] = arr[st_idx];
  for (long long int i = (1), i_end_ = (N); i < i_end_; i++) {
    for (long long int j = (0), j_end_ = (N); j < j_end_; j++) {
      if (dp[i][j] == INF) continue;
      chmin(dp[i + 1][j], dp[i][j] + arr[(j + st_idx) % N] + X);
      chmin(dp[i + 1][i], dp[i][j] + arr[(i + st_idx) % N]);
    }
  }
  for (long long int i = (0), i_end_ = (N); i < i_end_; i++) {
    chmin(minv, dp[N][i]);
  }
  cout << minv << '\n';
}
