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;
const int N = 2010;
long long a[N], dist[N];
priority_queue<pair<long long, int>, vector<pair<long long, int> >,
               greater<pair<long long, int> > >
    pq;
int main() {
  int n;
  long long x;
  scanf("%d%lld", &n, &x);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &a[i]);
    pq.push(make_pair(a[i], i));
  }
  while (!pq.empty()) {
    pair<long long, int> t = pq.top();
    pq.pop();
    long long pes = t.first;
    int at = t.second;
    int nxt = (at + n - 1) % n;
    if (a[nxt] <= pes + x) continue;
    a[nxt] = pes + x;
    dist[nxt] = dist[at] + 1;
    pq.push(make_pair(pes + x, nxt));
  }
  long long ans = 0;
  long long mx = 0;
  for (int i = 0; i < n; ++i) {
    mx = max(mx, dist[i]);
    ans += a[i] - dist[i] * 1ll * x;
  }
  ans += mx * 1ll * x;
  printf("%lld\n", ans);
}
