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;
signed main() {
  long long n, x;
  scanf("%lld %lld", &n, &x);
  ;
  vector<long long> data(n);
  for (long long i = 0; i < n; i++) {
    cin >> data[i];
  }
  vector<pair<long long, long long> > dp(
      n, pair<long long, long long>(-1, 1001000100010001000));
  for (long long i = 0; i < n; i++) {
    if (dp[i].second > data[i]) {
      dp[i].second = data[i];
      dp[i].first = i;
    } else {
      continue;
    }
    long long ptr = (i + 1) % n, cost = data[i] + x;
    while (ptr != i) {
      if (dp[ptr].second > cost) {
        dp[ptr].second = cost;
        dp[ptr].first = i;
      } else {
        break;
      }
      cost += x;
      ptr = (ptr + 1) % n;
    }
  }
  vector<long long> calc(n, 0);
  for (long long i = 0; i < n; i++) {
    calc[dp[i].first]++;
  }
  long long move = 0, ret = 0;
  for (long long i = 0; i < n; i++) {
    ret += calc[i] * data[i];
    move = max(move, calc[i]);
  }
  ret += (move - 1) * x;
  cout << ret << endl;
  return 0;
}
