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;
vector<pair<string, pair<int, int>>> vec;
void solve(long long N, long long x, vector<long long> a) {
  vector<vector<long long>> cost(N, vector<long long>(N));
  for (long long i = 0; i < N; i++) {
    for (long long j = 0; j < N; j++) {
      if (i <= j)
        cost[i][j] = a[i] + (j - i) * x;
      else
        cost[i][j] = a[i] + (j + N - i) * x;
    }
  }
  vector<long long> mincost(N, LLONG_MAX - 1);
  long long maxij = 0;
  for (long long i = 0; i < N; i++) {
    long long now = LLONG_MAX - 1;
    for (long long j = 0; j < N; j++) {
      if (mincost[i] > cost[j][i]) {
        if (i <= j)
          now = j - i;
        else
          now = j + N - i;
        mincost[i] = cost[j][i];
      } else if (mincost[i] == cost[j][i]) {
        if (i <= j)
          now = max(now, j - i);
        else
          now = max(now, j + N - i);
      }
    }
    maxij = max(maxij, now);
    mincost[i] -= x * now;
  }
  long long res = 0;
  for (long long i = 0; i < N; i++) {
    res += mincost[i];
  }
  res += x * maxij;
  cout << res << endl;
}
int main() {
  long long x;
  long long N;
  scanf("%lld", &N);
  vector<long long> a(N - 1 + 1);
  scanf("%lld", &x);
  for (int i = 0; i <= N - 1; i++) {
    scanf("%lld", &a[i]);
  }
  solve(N, x, a);
  return 0;
}
