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;
void input(int *N, int *x, int a[2000]) {
  cin >> *N >> *x;
  for (int i = 0; i < *N; i++) cin >> a[i];
}
void check(int N, int x, int a[2000]) {
  cout << N << " " << x << endl;
  for (int i = 0; i < N; i++) cout << a[i] << " ";
  cout << endl;
}
bool table_check(const vector<bool> &table, int count) {
  if (count < table.size()) return false;
  for (auto elem : table) {
    if (elem) return false;
  }
  return true;
}
int min_point(const vector<bool> &table, int a[2000]) {
  int MIN = 0;
  for (int i = 0; i < table.size(); i++)
    if (table[i]) MIN = i;
  for (int i = 0; i < table.size(); i++) {
    if (table[i]) MIN = (a[i] < a[MIN]) ? i : MIN;
  }
  return MIN;
}
bool point_check(int count, long long MIN_point, long long x,
                 vector<bool> &table) {
  if (count == 0) return true;
  if (x < MIN_point) {
    vector<int> buf;
    buf.clear();
    for (int i = 0; i < table.size(); i++) {
      if (table[i]) continue;
      buf.push_back(i);
    }
    for (int i = 0; i < buf.size(); i++) {
      table[buf[i]] = true;
    }
    for (int i = 0; i < buf.size(); i++) {
      table[(buf[i] == table.size() - 1) ? 0 : buf[i] + 1] = false;
    }
    return false;
  }
  return true;
}
int keiro(int N, int x, int a[2000]) {
  vector<bool> table;
  table.resize(N, true);
  int count = 0;
  int MIN = 0;
  int point = 0;
  while (1) {
    if (table_check(table, count)) break;
    MIN = min_point(table, a);
    if (point_check(count, a[MIN], x, table)) {
      table[MIN] = false;
      point += a[MIN];
    } else {
      point += x;
    }
    count++;
  }
  return point;
}
int main(int argc, char **argv) {
  int N = 0;
  int x = 0;
  int a[2000];
  input(&N, &x, a);
  cout << keiro(N, x, a) << endl;
}
