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 MAX = 1e18;
struct segK {
  int n;
  vector<pair<long long, long long> > dat;
  void init(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.resize(n * 2 - 1);
    for (int i = 0; i < n * 2 - 1; i++)
      dat[i] = pair<long long, long long>(MAX, 0);
  }
  void update(int k, long long p, long long c) {
    k += n - 1;
    dat[k] = pair<long long, long long>(p, c);
    while (k > 0) {
      k = (k - 1) / 2;
      if (dat[2 * k + 1].first < dat[2 * k + 2].first)
        dat[k] = dat[2 * k + 1];
      else if (dat[2 * k + 1].first > dat[2 * k + 2].first)
        dat[k] = dat[2 * k + 2];
      else if (dat[2 * k + 1].second < dat[2 * k + 2].second)
        dat[k] = dat[2 * k + 1];
      else
        dat[k] = dat[2 * k + 2];
    }
  }
  pair<long long, long long> query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return pair<long long, long long>(MAX, 0);
    if (l <= a && b <= r)
      return dat[k];
    else {
      pair<long long, long long> v1 = query(a, b, k * 2 + 1, l, (l + r) / 2);
      pair<long long, long long> v2 = query(a, b, k * 2 + 2, (l + r) / 2, r);
      if (v1.first > v2.first)
        return v2;
      else if (v1.first < v2.first)
        return v1;
      else if (v1.second < v2.second)
        return v1;
      else
        return v2;
    }
  }
};
int main() {
  int N;
  long long x;
  cin >> N >> x;
  vector<segK> segtree(N);
  vector<long long> a(N);
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    segtree[i].init(N);
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      segtree[i].update(j, a[(i + j) % N] + (N - j) % N * x,
                        (long long)(N - j) % N);
    }
  }
  long long ans = 0;
  vector<pair<long long, long long> > vec;
  long long cnt = 0;
  for (int i = 0; i < N; i++) {
    pair<long long, long long> temp = segtree[i].query(0, N, 0, 0, N);
    vec.push_back(temp);
    cnt = max(cnt, temp.second);
  }
  for (int i = 0; i < N; i++) ans += vec[i].first - x * vec[i].second;
  ans += cnt * x;
  cout << ans << endl;
}
