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 PII = pair<int, int>;
class Union_find {
 private:
  vector<int> par;
  vector<int> rank;
  int n;

 public:
  Union_find(int a) {
    n = a;
    for (int i = 0; i < n; i++) {
      par.push_back(i);
      rank.push_back(0);
    }
  }
  int find(int x) {
    if (par[x] == x) {
      return x;
    } else {
      return par[x] = find(par[x]);
    }
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (rank[x] < rank[y]) {
      par[x] = y;
    } else {
      par[y] = x;
      if (rank[x] == rank[y]) rank[x]++;
    }
  }
  bool same(int x, int y) { return find(x) == find(y); }
};
class SegTree {
  vector<ll> dat;
  int n;

 public:
  SegTree(int _n) {
    n = 1;
    while (n < _n) n *= 2;
    for (int i = 0; i < (n); i++) {
      dat.push_back(0);
    }
  }
};
int main(void) {
  ll N, x;
  ll res = 1e18;
  ll a[2002];
  cin >> N >> x;
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
  }
  for (int i = 0; i < (N); i++) {
    ll cnt = 0;
    for (int j = 0; j < (N); j++) {
      ll min_a = 1e18;
      for (int k = 0; k < (i + 1); k++) {
        min_a = min(min_a, a[(j - k + N) % N]);
      }
      cnt += min_a;
    }
    cnt += x * i;
    res = min(res, cnt);
  }
  cout << res << endl;
}
