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;
int cc = 0;
int N;
long long int x;
long long int a[2000];
bool member(int i, queue<int> used) {
  bool tmp = false;
  queue<int> sub = used;
  while (!sub.empty()) {
    if (i == sub.front()) {
      tmp = true;
      break;
    }
    sub.pop();
  }
  return tmp;
}
int min_array(long long int a[], queue<int> used) {
  int ans = -1;
  long long int min = 1000000001;
  for (int i = 0; i < N; i++) {
    if (!member(i, used)) {
      if (a[i] < min) {
        min = a[i];
        ans = i;
      }
    }
  }
  return ans;
}
queue<int> next(queue<int> used) {
  queue<int> ans, sub = used;
  while (!sub.empty()) {
    if (sub.front() != N - 1) {
      ans.push(sub.front() + 1);
    } else {
      ans.push(0);
    }
    sub.pop();
  }
  cc++;
  return ans;
}
queue<int> add(queue<int> used, int i) {
  queue<int> ans = used;
  ans.push(i);
  cc++;
  return ans;
}
long long int min(long long int a, long long int b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
long long int ans(queue<int> used) {
  if (cc > 2000) return 10000000000;
  if (used.empty()) {
    int i = min_array(a, used);
    used.push(i);
    return a[i] + ans(used);
  } else {
    int i = min_array(a, used);
    if (i == -1) {
      return 0;
    } else {
      return min((x + ans(next(used))), (a[i] + ans(add(used, i))));
    }
  }
}
int main() {
  scanf("%d %lld", &N, &x);
  for (int i = 0; i < N; i++) {
    scanf("%lld", &a[i]);
  }
  queue<int> used;
  printf("%lld", ans(used) % 10000000000);
  return 0;
}
