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>
#pragma GCC optimize("Ofast")
#pragma GCC target("tune=native")
#pragma GCC target("avx")
using namespace std;
const long long mod = 1000000007;
long long n, x, a[2010], ret;
vector<pair<long long, long long> > operation;
signed main() {
  scanf("%lld", &(n));
  scanf("%lld", &(x));
  operation.resize(n);
  for (long long i = 0; i < (long long)(n); i++) scanf("%lld", &(a[i]));
  for (long long i = 0; i < (long long)(n); i++) operation[i].second = a[i];
  for (long long _ = 0; _ < (long long)(n); _++) {
    for (long long i = 0; i < (long long)(n); i++) {
      if (a[(i + 1) % n] > a[i] + x) {
        a[(i + 1) % n] = a[i] + x;
        operation[(i + 1) % n].first = operation[i].first + 1;
        operation[(i + 1) % n].second = a[(i - operation[i].first + n) % n];
      }
    }
  }
  long long maxcast = 0;
  for (long long i = 0; i < (long long)(n); i++) {
    ret += operation[i].second;
    maxcast = max(maxcast, operation[i].first);
  }
  cout << ret + maxcast * x << endl;
}
