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;
long long a[2000];
pair<int, long long> d[2000];
inline int dist(int s, int e, int n) {
  if (s <= e) {
    return e - s;
  } else {
    return n + e - s;
  }
}
int main() {
  int N;
  long long x;
  cin >> N >> x;
  int min_i = 0;
  long long min = 1000000000000000;
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
    if (min > a[i]) {
      min = a[i];
      min_i = i;
    }
    d[i] = make_pair(i, a[i]);
  }
  d[min_i] = make_pair(min_i, min);
  int prev = min_i;
  int ind = (min_i == N - 1) ? 0 : min_i + 1;
  while (true) {
    if (ind == min_i) {
      break;
    }
    if (d[ind].second > d[prev].second + x) {
      d[ind] = make_pair(d[prev].first, d[prev].second + x);
    }
    prev = ind;
    ind = (ind == N - 1) ? 0 : ind + 1;
  }
  long long cost = 0;
  int max_d = 0;
  for (int i = 0; i < (N); i++) {
    cost += a[d[i].first];
    if (max_d < dist(d[i].first, i, N)) {
      max_d = dist(d[i].first, i, N);
    }
  }
  cout << cost + max_d * x << endl;
  return 0;
}
