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;
long long c[4000];
signed main() {
  long long a, b;
  scanf("%lld%lld", &a, &b);
  for (long long d = 0; d < a; d++) {
    scanf("%lld", &c[d]);
    c[d + a] = c[d];
  }
  long long MIN = LLONG_MAX;
  for (long long i = 0; i < a; i++) {
    long long sum = b * i;
    deque<pair<long long, long long> > Q;
    long long k = LLONG_MAX;
    for (long long j = 0; j <= i; j++) {
      while (Q.size() && Q.back().second >= c[j]) Q.pop_back();
      Q.push_back({j, c[j]});
      k = min(k, c[j]);
    }
    sum += k;
    for (long long j = 1; j < a; j++) {
      if (Q.front().first == j - 1) Q.pop_front();
      while (Q.size() && Q.back().second >= c[i + j]) Q.pop_back();
      Q.push_back({i + j, c[i + j]});
      sum += Q.front().second;
    }
    MIN = min(MIN, sum);
  }
  printf("%lld\n", MIN);
}
