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")
using namespace std;
const long double PI = acos(-1.0L);
const long long MOD = 1000000007LL;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long N, x;
  cin >> N >> x;
  vector<long long> A(N);
  for (int i = 0; i < (int)(N); i++) cin >> A[i];
  vector<long long> B(N, 1e18), cnt(N, 1e18);
  for (int i = 0; i < (int)(N); i++) {
    long long val = 1e18;
    for (int j = 0; j < (int)(N); j++) {
      int k = i - j;
      if (k < 0) k += N;
      if (val >= A[j] + k * x) {
        if (val == A[j] + k * x && A[j] > B[i]) continue;
        val = A[j] + k * x;
        B[i] = A[j];
        cnt[i] = k;
      }
    }
  }
  cout << accumulate(B.begin(), B.end(), 0LL) +
              *max_element(cnt.begin(), cnt.end()) * x
       << endl;
}
