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;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
  ll N, x;
  ll ans = 100000000000;
  cin >> N >> x;
  vector<ll> a(N);
  vector<pair<ll, ll>> b(N, {100000000000, 0});
  vector<vector<pair<ll, ll>>> table(N, vector<pair<ll, ll>>(N));
  for (int i = 0; i < N; i++) cin >> a[i];
  for (int i = 0; i < N; i++) {
    ll res = 0;
    for (int j = 0; j < N; j++) {
      if (i == 0)
        b[j] = {a[j], 0};
      else if (j - i < 0) {
        if (b[j].first > a[N - i + j] + i * x) b[j] = {a[N - i + j] + i * x, i};
      } else if (j - i >= 0) {
        if (b[j].first > a[j - i] + i * x) b[j] = {a[j - i] + i * x, i};
      }
    }
    for (int k = 0; k < N; k++) {
      res += b[k].first - b[k].second * x;
    }
    res += i * x;
    ans = min(ans, res);
  }
  cout << ans << endl;
}
