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 P = pair<long long, long long>;
const long long INF = (long long)1e9;
const LL INFL = (LL)1e15;
const long long MOD = 1e9 + 7;
long long dy[] = {0, 0, 1, -1, 0};
long long dx[] = {1, -1, 0, 0, 0};
long long n, x;
vector<long long> a;
long long dp[2005];
set<P> st;
long long cnt[2005];
long long pre_color(long long i) { return (i != 0 ? i - 1 : n - 1); }
long long next_color(long long i) { return (i + 1) % n; }
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> x;
  a.resize(n);
  for (long long i = (long long)(0); i < (long long)(n); ++i) cin >> a[i];
  for (long long i = (long long)(0); i < (long long)(n); ++i) dp[i] = a[i];
  for (long long i = (long long)(0); i < (long long)(n); ++i) cnt[i] = 1;
  for (long long i = (long long)(0); i < (long long)(n); ++i) {
    if (cnt[i] == 1) {
      for (long long j = (long long)(0); j < (long long)(n - 1); ++j) {
        if (dp[i] * (j + 1) + x * (j + 1) <= dp[next_color(i + j)]) {
          cnt[i]++;
          cnt[next_color(i + j)] = 0;
          dp[next_color(i + j)] = dp[i] * (j + 1) + x * (j + 1);
        } else
          break;
      }
    }
  }
  long long ans = 0;
  for (long long i = (long long)(0); i < (long long)(n); ++i)
    ans += a[i] * cnt[i];
  long long max_cnt = 0;
  for (long long i = (long long)(0); i < (long long)(n); ++i)
    max_cnt = max(max_cnt, cnt[i]);
  ans += (max_cnt - 1) * (x);
  cout << ans << endl;
}
