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>
const int INF = 1e9;
const int MO = 1e9 + 7;
using LL = long long;
const LL LINF = 1e18;
const double EPS = 1e-10;
using namespace std;
class PalseError {};
class Edge {
 public:
  LL from, to, value;
  Edge(LL a, LL b, LL c) {
    from = a;
    to = b;
    value = c;
  }
  Edge(LL a, LL b) {
    from = a;
    to = b;
  }
};
class SegTreeRMQ {
 public:
  SegTreeRMQ(int N) { init(N); }
  void init(int N) {
    int second = 1;
    while (second < N) {
      second *= 2;
    }
    dat = std::vector<LL>(second * 2, 1e9);
    n = second;
  }
  void update(int i, int x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = min(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  int query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return 1e9;
    if (a <= l && r <= b)
      return dat[k];
    else {
      int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return min(vl, vr);
    }
  }
  int getSize() { return n; }
  void show() {
    for (int a = 0; a < n; a++) {
      cout << dat[a] << " ";
    }
    cout << endl;
    return;
  }

 private:
  int n;
  std::vector<LL> dat;
};
int main() {
  int N, x;
  cin >> N >> x;
  SegTreeRMQ SegTree(N * 2);
  vector<LL> vec;
  for (int a = 0; a < N; a++) {
    int b;
    cin >> b;
    vec.push_back(b);
    SegTree.update(a, b);
  }
  for (int a = 0; a < N; a++) {
    vec.push_back(vec.at(a));
    SegTree.update(a + N, vec.at(a));
  }
  LL ans = 1e18;
  for (int b = 0; b <= N; b++) {
    LL ins = b * x;
    for (int a = N; a < 2 * N; a++) {
      ins += SegTree.query(a - b, a + 1, 0, 0, SegTree.getSize());
    }
    ans = min(ans, ins);
  }
  cout << ans << endl;
}
