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;
using LL = long long;
using ULL = unsigned long long;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
bool b[2000 + 10];
int p[2000 + 10];
void solve(long long N, long long x, vector<long long> a) {
  for (int i = 0; i < N; ++i) {
    p[i] = i;
  }
  sort(p + 1, p + 1 + N, [&](int x, int y) { return a[x] < a[y]; });
  LL ans = INT_MAX * 1000LL;
  for (int i = 0; i < N; ++i) {
    memset(b, true, sizeof(b));
    LL nowans = 0;
    for (int j = 1; j <= N; ++j) {
      int now = p[j];
      for (int k = 1; k <= i + 1; ++k) {
        if (!b[now]) break;
        b[now] = false;
        nowans += a[p[j]];
        ++now;
        if (now >= N) now = 0;
      }
    }
    ans = min(ans, nowans + i * x);
  }
  cout << ans << endl;
}
int main() {
  long long N;
  long long x;
  scanf("%lld", &N);
  scanf("%lld", &x);
  vector<long long> a(N - 1 + 1);
  for (int i = 0; i <= N - 1; i++) {
    scanf("%lld", &a[i]);
  }
  solve(N, x, a);
  return 0;
}
