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 inf = 1 << 28;
const long long INF = 1ll << 50;
const double pi = acos(-1);
const double eps = 1e-8;
const int mod = 1e9 + 7;
const vector<int> emp;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int DX[8] = {-1, -1, -1, 0, 0, 1, 1, 1},
          DY[8] = {1, 0, -1, 1, -1, 1, 0, -1};
class RMQ {
 private:
  int n;
  vector<int> rmq;
  long long Queryfunc(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return inf;
    if (a <= l && r <= b)
      return rmq[k];
    else {
      int mid = (l + r) / 2;
      int vector<long long> = Queryfunc(a, b, k * 2 + 1, l, mid);
      int vr = Queryfunc(a, b, k * 2 + 2, mid, r);
      return (vector<long long> < vr) ? vector<long long> : vr;
    }
  }

 public:
  RMQ(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    rmq = vector<int>(2 * n - 1);
  }
  void Update(int k, int x) {
    k += n - 1;
    rmq[k] = x;
    while (k > 0) {
      k = (k - 1) / 2;
      rmq[k] = min(rmq[k * 2 + 1], rmq[k * 2 + 2]);
    }
  }
  long long Query(int a, int b) { return Queryfunc(a, b, 0, 0, n); }
};
int n, x;
int main() {
  cin >> n >> x;
  RMQ r(2 * n);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    r.Update(i, a);
    r.Update(i + n, a);
  }
  long long ans = inf;
  for (int i = 0; i < n; i++) {
    long long res = i * x;
    for (int j = 0; j < n; j++) res += r.Query(j, j + i + 1);
    ans = min(ans, res);
  }
  cout << ans << endl;
}
