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 int MAXN = 2010;
const int MAXLOG = 11;
const int INF = 0x3f3f3f3f;
using namespace std;
int N, X, minv[MAXLOG][MAXN], log2[MAXN];
inline int rmq(int l, int r) {
  int len = log2[r - l + 1];
  return min(minv[len][l], minv[len][r - (1 << len) + 1]);
}
inline int radq(int x, int d) {
  int ret, l = x - d;
  ret = rmq(max(1, l), x);
  if (l < 1) ret = min(ret, rmq(N - (1 - l) + 1, N));
  return ret;
}
int main() {
  int i;
  scanf("%d%d", &N, &X);
  for (i = 1; i <= N; i++) scanf("%d", &minv[0][i]);
  log2[1] = 0;
  for (i = 2; i <= N; i++) log2[i] = log2[i >> 1] + 1;
  for (int k = 1; (1 << k) <= N; k++)
    for (i = 1; i + (1 << k) - 1 <= N; i++)
      minv[k][i] = min(minv[k - 1][i], minv[k - 1][i + (1 << (k - 1))]);
  long long Ans = INF;
  for (int r = 0; r <= N; r++) {
    long long CurAns = (long long)X * r;
    for (i = 1; i <= N; i++) CurAns += radq(i, r);
    Ans = min(Ans, CurAns);
  }
  printf("%lld", Ans);
  return 0;
}
