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 N;
int x;
int a[2000];
int b[2000];
bool magiced[2000];
int get_min_index_of_a() {
  int index = 0;
  int min = 1000000000;
  for (int i = 0; i < N; i++) {
    if (a[i] < min) {
      min = a[i];
      index = i;
    }
  }
  return index;
}
int get_max_of_b() {
  int max = 0;
  for (int i = 0; i < N; i++) {
    if (b[i] > max) {
      max = b[i];
    }
  }
  return max;
}
void solve() {
  int min_index = get_min_index_of_a();
  if (min_index == 0) {
    memcpy(b, a, sizeof(int) * N);
  } else {
    int b_pointer = 0;
    for (int i = min_index; i < N; i++) {
      b[b_pointer] = a[i];
      b_pointer++;
    }
    for (int i = 0; i < min_index; i++) {
      b[b_pointer] = a[i];
      b_pointer++;
    }
  }
  int magic_time = 0;
  while (1) {
    bool changed = false;
    for (int i = N - 1; i >= 1; i--) {
      if (b[i] - b[i - 1] > x) {
        changed = true;
        magiced[i] = true;
        b[i] = b[i - 1] + x;
      }
    }
    if (!changed) {
      break;
    }
    magic_time++;
  }
  long long wait_time = 0;
  for (int i = 0; i < N; i++) {
    if (!magiced[i]) {
      int i_times = 1;
      for (int j = i + 1; j < N; j++) {
        if (!magiced[j]) break;
        i_times++;
      }
      wait_time += b[i] * i_times;
    }
  }
  long long ans = wait_time + magic_time * x;
  cout << ans << endl;
}
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
  }
  solve();
  return 0;
}
