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 INF = 1e9;
const long long LINF = 1e18;
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long N, x;
  cin >> N >> x;
  vector<long long> a(N);
  for (auto &in : a) cin >> in;
  vector<long long> cusum(2 * N + 1, 0);
  for (int i = 0; i < 2 * N; i++) {
    cusum[i + 1] = cusum[i] + a[i % N];
  }
  vector<pair<long long, long long> > ps(N);
  for (int i = 0; i < N; i++) ps[i] = pair<long long, long long>(a[i], i + 1);
  sort(ps.begin(), ps.end());
  long long ans = 0;
  vector<int> keep(N + 1, 0);
  for (int i = 0; i < N; i++) {
    long long F = ps[i].first;
    long long l = ps[i].second, r = ps[i].second;
    if (keep[l] % N == 1) continue;
    for (; r < l + N;) {
      if (keep[r % N] == 1) break;
      long long n = r - (l - 1);
      if (cusum[r] - cusum[l - 1] >= F * n + x * (n - 1)) {
        r++;
      } else {
        break;
      }
    }
    long long n = r - l;
    ans += F * n + x * (n - 1);
    for (long long j = l; j < r; j++) {
      long long k = j;
      if (k > N) k -= N;
      keep[k] = 1;
    }
  }
  cout << ans << endl;
  return 0;
}
