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;
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
  std::fill((T *)array, (T *)(array + N), val);
}
pair<long long int, long long int> dist[3000];
long long int v[3000];
int main() {
  long long int n, x;
  cin >> n >> x;
  Fill(v, 1e10);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long int maxIndex = 0;
  for (int i = 0; i < n; i++) {
    long long int minDist = v[i];
    long long int minIndex = 0;
    long long int from = i;
    for (int j = i - 1; j > i - n; j--) {
      long long int mj = (j + n) % n;
      long long int d = v[mj] + (i - j) * x;
      if (minDist > d) {
        minDist = d;
        minIndex = i - j;
        from = mj;
      }
    }
    dist[i] = make_pair(from, minIndex);
    maxIndex = max(maxIndex, minIndex);
  }
  long long int sum = 0;
  for (int i = 0; i < n; i++) {
    sum += v[dist[i].first];
  }
  sum += maxIndex * x;
  cout << sum << endl;
  return 0;
}
