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>
const long long INFN = 1LL << 61;
const int MAXN = 2016;
long long dp[MAXN][MAXN][2]{};
int array[MAXN]{};
long long Ans, K, tem;
int N;
inline long long getMinAr(long long A, long long B) {
  return (array[A] < array[B]) ? A : B;
}
int main() {
  scanf(" %d%lld", &N, &K);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &array[i]);
    for (int j = 0; j <= N; ++j) dp[i][j][1] = INFN;
  }
  array[0] = (int)1E9 + 5;
  for (int i = 1; i <= N; ++i) {
    for (int j = 1; j <= N; ++j)
      if (i != j) {
        if (j < i) {
          long long next = (array[j] + K * (i - j));
          if (next <= dp[i][i - j][1]) {
            if (next == dp[i][i - j][1])
              dp[i][i - j][0] = getMinAr(j, dp[i][i - j][0]);
            else
              dp[i][i - j][0] = j;
            dp[i][i - j][1] = next;
          }
        }
        if (j > i) {
          long long next = (array[j] + K * (N - j + i));
          if (next <= dp[i][N - j + i][1]) {
            if (next == dp[i][N - j + i][1])
              dp[i][N - j + i][0] = getMinAr(j, dp[i][N - j + i][0]);
            else
              dp[i][N - j + i][0] = j;
            dp[i][N - j + i][1] = next;
          }
        }
      }
    dp[i][0][0] = i;
    dp[i][0][1] = array[i];
  }
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= N; ++j)
      if (dp[i][j - 1][1] <= dp[i][j][1]) {
        if (dp[i][j - 1][1] == dp[i][j][1])
          dp[i][j][0] = getMinAr(dp[i][j - 1][0], dp[i][j][0]);
        else
          dp[i][j][0] = dp[i][j - 1][0];
        dp[i][j][1] = dp[i][j - 1][1];
      }
  Ans = INFN;
  for (int j = 0; j <= N; ++j) {
    tem = K * j;
    for (int i = 1; i <= N; ++i) tem += array[dp[i][j][0]];
    Ans = std::min(tem, Ans);
  }
  printf("%lld\n", Ans);
  return 0;
}
