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 long long INF = 1e15;
const long long MAXN = 2011;
long long n, x, t[MAXN * 8], a[MAXN * 2];
void build(long long v = 1, long long tl = 1, long long tr = n) {
  if (tl == tr) {
    t[v] = a[tl];
    return;
  }
  long long tm = (tl + tr) >> 1;
  build(v + v, tl, tm);
  build(v + v + 1, tm + 1, tr);
  t[v] = min(t[v + v], t[v + v + 1]);
}
long long get(long long l, long long r, long long v = 1, long long tl = 1,
              long long tr = n) {
  if (r < tl || tr < l) {
    return INF;
  }
  if (l <= tl && tr <= r) {
    return t[v];
  }
  long long tm = (tl + tr) >> 1;
  return min(get(l, r, v + v, tl, tm), get(l, r, v + v + 1, tm + 1, tr));
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> x;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    ans += a[i];
  }
  n *= 2;
  for (int i = (n / 2) + 1, j = 1; i <= n; ++i, ++j) {
    a[i] = a[j];
  }
  build();
  for (int len = 1; len < n; ++len) {
    long long sum = 0;
    for (int l = 2, r = 2 + len; r <= n; ++l, ++r) {
      sum += get(l, r);
    }
    ans = min(ans, sum + (len * x));
  }
  cout << ans;
  return 0;
}
