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;
#pragma comment(linker, "/STACK:64000000")
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const long long maxn = 2001;
const long long inf = 1000000007;
const long long mod = 1000000007;
long long n, a[maxn], sum, x, k[maxn], su[maxn];
set<pair<long long, long long> > s;
long long dst(long long y, long long z) {
  if (z > y)
    return z - y;
  else
    return z + n - y;
}
void solve() {
  while (!s.empty()) {
    long long ver = s.begin()->second;
    long long val = s.begin()->first;
    s.erase(s.begin());
    if (val != a[ver]) continue;
    for (int i = 0; i < n; i++) {
      if (i == ver) continue;
      if (a[i] > a[ver] + x * dst(ver, i)) {
        a[i] = a[ver] + x * dst(ver, i);
        k[i] = dst(ver, i);
        su[i] = a[ver];
        s.insert(make_pair(a[i], i));
      }
    }
  }
}
int main() {
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    su[i] = a[i];
    s.insert(make_pair(a[i], i));
  }
  solve();
  long long ms = 0;
  for (int i = 0; i < n; i++) {
    ms = max(ms, k[i]);
    sum += su[i];
  }
  cout << sum + x * ms;
  return 0;
}
