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 secNeed(vector<int> &a, int ori, int moveStep, int N, int x) {
  int index = ori >= moveStep ? ori - moveStep : ori + N - moveStep;
  return a[index] + moveStep * x;
}
int main() {
  int N, x;
  int maxStep = 0;
  cin >> N >> x;
  vector<int> a(N);
  vector<vector<pair<int, int>>> minTime(N);
  for (int i = 0; i < N; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < N; ++i) {
    minTime[i].push_back({i, 0});
    for (int moveStep = 1; moveStep < N; ++moveStep) {
      long long pre = secNeed(a, minTime[i][moveStep - 1].first,
                              minTime[i][moveStep - 1].second, N, x);
      long long cur = secNeed(a, i, moveStep, N, x);
      if (cur < pre) {
        minTime[i].push_back({i, moveStep});
        if (moveStep > maxStep) maxStep = moveStep;
      } else {
        minTime[i].push_back(minTime[i][moveStep - 1]);
      }
    }
  }
  long long minTTTime = 9223372036854775807;
  for (long long i = 0; i <= maxStep; ++i) {
    long long sum = 0;
    for (int j = 0; j < N; ++j) {
      int index = minTime[j][i].first - minTime[j][i].second;
      if (index < 0) index += N;
      sum += a[index];
    }
    sum += i * (long long)x;
    if (sum < minTTTime) minTTTime = sum;
  }
  cout << minTTTime << endl;
  return 0;
}
