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;
vector<long long int> v;
vector<long long int> b;
long long int target, n, x, i;
long long int cost = 0;
long long int sum = 0;
long long int merit() {
  long long int c = 0;
  long long int next;
  for (long long int i = 0; i < b.size(); ++i) {
    if (b[i] >= v.size()) {
      next = 0;
    } else {
      next = b[i] - 1 + 1;
    }
    c += v[next] - v[b[i] - 1];
  }
  return c - x;
}
void m() {
  for (long long int i = 0; i < b.size(); ++i) {
    if (b[i] - 1 >= v.size() - 1) {
      b[i] = 1;
    } else {
      b[i]++;
    }
  }
}
bool find(long long int t) {
  for (auto v : b) {
    if (v == t) {
      return true;
    }
  }
  return false;
}
int min_index() {
  long long int ti = 0;
  long long int m = 1000000000;
  for (long long int i = 0; i < v.size(); i++) {
    if (m > v[i] && !find(i + 1)) {
      m = v[i];
      ti = i;
    }
  }
  return ti;
}
void showb() {
  for (auto i : b) {
    cout << i << " ";
  }
  cout << endl;
}
int main() {
  cin >> n >> x;
  target = n;
  while (n--) {
    long long int a;
    cin >> a;
    v.push_back(a);
  }
  while (b.size() < target) {
    if (merit() < 0) {
      i = min_index();
      b.push_back(i + 1);
      cost += v[i];
    } else {
      m();
      cost += x;
    }
  }
  cout << cost << endl;
  return 0;
}
