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 = 1000000001;
const long long INF = 1e18 * 4;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dx2[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy2[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long MAX_N = 1 << 13;
int n;
vector<long long> dat(2 * MAX_N - 1);
void init(int n_) {
  n = 1;
  while (n < n_) n *= 2;
  for (int i = 0; i < 2 * n - 1; i++) dat[i] = inf;
}
void update(int k, int a) {
  k += n - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
  }
}
int query(int a, int b, int k, int l, int r) {
  if (r <= a || b <= l) {
    return inf;
  }
  if (a <= l && r <= b) {
    return dat[k];
  } else {
    long long ul = query(a, b, k * 2 + 1, l, (l + r) / 2);
    long long ur = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return min(ul, ur);
  }
}
int main() {
  int _n, x, i, j;
  scanf("%d%d", &_n, &x);
  vector<int> a(_n);
  for (i = 0; i < _n; i++) {
    scanf("%d", &a[i]);
  }
  init(2 * _n);
  for (i = 0; i < _n; i++) {
    update(i, a[i]);
    update(i + _n, a[i]);
  }
  long long ans = INF;
  for (i = 0; i < _n; i++) {
    long long res = 0;
    res += i * x;
    for (j = 0; j < _n; j++) {
      res += query(j, j + i + 1, 0, 0, n);
    }
    ans = min(ans, res);
  }
  for (i = _n; i < 2 * _n; i++) {
    long long res = 0;
    res += i * x;
    for (j = 0; j < _n; j++) {
      res += query(j - i, j + 1, 0, 0, n);
    }
    ans = min(ans, res);
  }
  printf("%lld\n", ans);
}
