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;
const int N = 2e3 + 7;
const long long INF = 1e18 + 7;
long long n, x;
long long a[N];
long long dp[N][N];
vector<int> p[N];
int main() {
  scanf("%lld%lld", &n, &x);
  for (int i = 0; i < int(n); i++) scanf("%lld", &a[i]);
  for (int i = 0; i < int(n); i++) {
    long long step = 0;
    for (int j = i + 1 - 1; j >= int(0); j--) {
      dp[i][step] = a[j] + step * x;
      step++;
    }
    for (int j = n - 1; j >= int(i + 1); j--) {
      dp[i][step] = a[j] + step * x;
      step++;
    }
  }
  long long res = 0;
  for (int i = 0; i < int(n); i++) {
    long long cur = INF;
    long long pos = 0;
    long long step = 0;
    for (int j = 0; j < int(n); j++) {
      if (cur > dp[i][j]) {
        cur = dp[i][j];
        pos = (i - j + n) % n;
        step = j;
      }
    }
    p[step].push_back(pos);
  }
  int de = 0;
  for (int i = 0; i < int(n); i++) {
    if (((int)(p[i]).size())) {
      de = i;
    }
    for (auto j : p[i]) {
      res += a[j];
    }
  }
  res += de * x;
  printf("%lld\n", res);
  return 0;
}
