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;
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long N, x;
  cin >> N >> x;
  vector<long long> A(N);
  vector<long long> dp(N);
  for (long long i = 0; i < N; i++) {
    dp[i] = 9223372036854775807;
  }
  for (long long i = 0; i < N; i++) cin >> A[i];
  for (long long i = 0; i < N; i++) {
    for (long long j = (i + 1); j < (N); ++j) {
      dp[i] = min(dp[i], A[j] + x * (N - j - 1) + x * (i + 1));
    }
  }
  dp[0] = min(dp[0], A[0]);
  long long a = 0;
  for (long long i = (1); i < (N); ++i) {
    if (A[i] < dp[a] + x * (i - a)) {
      a = i;
      dp[i] = min(dp[i], A[i]);
    } else
      dp[i] = min(dp[i], dp[a] + x * (i - a));
  }
  long long cnt = 1;
  long long ans = 0;
  a = 0;
  long long i = 0;
  bool F[2000] = {false};
  if (dp[0] > dp[N - 1]) {
    while (dp[i] < A[i]) {
      cnt++;
      F[i] = true;
      i++;
    }
    a = i + 1;
    i = N - 1;
    F[i] = true;
    while (dp[i] < A[i]) {
      cnt++;
      i--;
      F[i] = true;
    }
    ans += A[i] * cnt + x * (cnt - 1);
  }
  cnt = 1;
  for (long long i = 0; i < N; i++) {
    if (F[i]) {
      a = i + 1;
      continue;
    }
    if (i == N - 1) {
      ans += (cnt - 1) * x + A[a] * cnt;
      cnt = 0;
      a = i + 1;
      break;
    }
    if (dp[i + 1] == A[i + 1]) {
      ans += (cnt - 1) * x + A[a] * cnt;
      cnt = 1;
      a = i + 1;
    } else {
      cnt++;
    }
  }
  cout << ans << endl;
  return 0;
}
