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;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class A, size_t N, class T>
void Fill(A (&a)[N], const T &v) {
  fill((T *)a, (T *)(a + N), v);
}
const int INF = 0x3fffffff;
int main() {
  long long int N, x, mx = 0;
  long long int a[2005], c[2005], cx[2005] = {};
  Fill(c, INF * INF);
  cin >> N >> x;
  for (int i = 0; i < (N); i++) cin >> a[i];
  for (int i = 0; i < (N); i++)
    for (int j = 0; j < (N); j++) {
      if (chmin(c[(i + j) % N], a[i] + x * (long long int)j)) {
        cx[(i + j) % N] = (long long int)j;
      }
    }
  for (int i = 0; i < (N); i++) chmax(mx, cx[i]);
  long long int ans = mx * x;
  for (int i = 0; i < (N); i++) ans += c[i] - cx[i] * x;
  cout << ans << endl;
  return 0;
}
