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>
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
int main(void) {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  pair<long long, long long> dp[n];
  priority_queue<pair<long long, int>, vector<pair<long long, int>>,
                 greater<pair<long long, int>>>
      que;
  for (int i = 0; i < n; i++) {
    dp[i] = {a[i], 0};
    que.push({a[i], i});
  }
  while (que.size()) {
    auto p = que.top();
    que.pop();
    long long tmp = p.first;
    int id = p.second;
    if (tmp != dp[id].first) continue;
    if (chmin(dp[(id + 1) % n].first, tmp + x)) {
      que.push({tmp + x, (id + 1) % n});
      dp[(id + 1) % n].second++;
    }
  }
  long long res = 0, mag = 0;
  for (int i = 0; i < n; i++) {
    res += dp[i].first - x * dp[i].second;
    chmax(mag, dp[i].second);
  }
  res += mag * x;
  cout << res << endl;
}
