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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
pair<int, int> a[2002];
int b[2002], vis[2002];
int main() {
  int i, j, k, ca = 0, n, m, K;
  scanf("%d%d", &n, &K);
  for (int i = 0; i < n; i++)
    scanf("%d", &a[i].first), a[i].second = i, b[i] = a[i].first;
  sort(a, a + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) ans += a[i].first;
  for (int k = 1; k < n; k++) {
    i = 0;
    if (1LL * k * K >= ans) break;
    for (int j = 0; j < n; j++) vis[j] = 0;
    long long sum = 1LL * k * K;
    for (int j = 0; j < n; j++)
      if (vis[a[j].second] == 0) {
        int second = a[j].second, val = a[j].first;
        vis[second] = 1;
        sum += val;
        for (int l = 1; l < k + 1; l++) {
          int first = (second + l) % n;
          if (b[first] > b[second]) sum += val, vis[first] = 1;
          if (sum >= ans) break;
        }
      }
    Min(ans, sum);
  }
  printf("%lld\n", ans);
  return 0;
}
