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;
long long A[2000], B[2000];
long long dp[2][2001];
long long f(long long b) {
  for (long long j = 0; j <= b; j++) {
    dp[0][j] = (1LL << 60);
    dp[1][j] = (1LL << 60);
  }
  dp[0][0] = 0;
  long long lmin = 0;
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j <= b; j++) dp[1][j] = (1LL << 60);
    dp[1][1] = lmin + B[i];
    lmin = dp[1][1];
    for (long long j = 1; j < b && j <= i; j++) {
      dp[1][j + 1] = dp[0][j] + B[i - j];
      lmin = min(lmin, dp[1][j + 1]);
    }
    swap(dp[0], dp[1]);
  }
  return X * (b - 1) + *min_element(dp[0], dp[0] + b + 1);
}
signed main() {
  cin >> N >> X;
  for (long long i = 0; i < N; i++) cin >> A[i];
  long long mi = min_element(A, A + N) - A;
  for (long long i = 0; i < N; i++) B[i] = A[(mi + i) % N];
  long long m = (1LL << 60);
  for (long long i = 1; i <= N; i++) m = min(m, f(i));
  cout << m << "\n";
  return 0;
}
