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 double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
long long n, a[MAXN], x;
long long pre[MAXN], cost[MAXN], com[MAXN];
long long dp[MAXN][50];
void init() {
  for (int i = 1; i <= n; i++) dp[i][0] = a[i];
  int nlog = (int)log2(n);
  for (int j = 1; j <= nlog; j++)
    for (int i = 1; i <= n; i++)
      if ((i + (1 << j) - 1) <= n)
        dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
long long rmq(int s, int e) {
  if (s > e) return INF;
  int nlog = (int)log2(e - s + 1);
  return min(dp[s][nlog], dp[e - (1 << nlog) + 1][nlog]);
}
int main() {
  cin >> n >> x;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  init();
  long long ans = INF;
  for (int k = 0; k < n; k++) {
    for (int i = 1; i <= n; i++) {
      cost[i] = a[i];
      if (i > 1 && k) {
        int L = max(1, i - k);
        int R = i - 1;
        cost[i] = min(cost[i], rmq(L, R));
        if (i - k < 1) {
          int L = n + 1 - (k - (i - 1));
          int R = n;
          cost[i] = min(cost[i], rmq(L, R));
        }
      } else if (k) {
        int L = n + 1 - k;
        int R = n;
        cost[i] = min(cost[i], rmq(L, R));
      }
    }
    long long sum = k * x;
    for (int i = 1; i <= n; i++) sum += cost[i];
    ans = min(sum, ans);
  }
  printf("%lld\n", ans);
  return 0;
}
