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;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
void solve(long long N, long long x, std::vector<long long> a) {
  vector<long long> orig(N);
  long long pre = 0;
  orig[0] = pre;
  long long t = 1;
  for (int(n) = 0; (n) < (2 * N - 1); ++(n)) {
    if (n == 0) {
      continue;
    }
    long long i = n % N;
    if (a[pre] + x * t > a[i]) {
      orig[i] = i;
      pre = i;
      t = 1;
    } else {
      orig[i] = pre;
      t++;
    }
  }
  long long sum = 0;
  for (int(n) = 0; (n) < (N); ++(n)) {
    sum += a[orig[n]];
  }
  long long maxT = 0;
  long long tmpT = 0;
  pre = 0;
  for (int(n) = 0; (n) < (2 * N - 1); ++(n)) {
    if (n == 0) {
      continue;
    }
    long long i = n % N;
    if (pre == orig[i]) {
      tmpT++;
      if (tmpT == N - 1) {
        maxT = tmpT;
        break;
      }
    } else {
      tmpT = 0;
      pre = orig[i];
      maxT = max(maxT, tmpT);
    }
  }
  cout << sum + x * maxT << endl;
}
int main() {
  long long N;
  scanf("%lld", &N);
  long long x;
  scanf("%lld", &x);
  std::vector<long long> a(N);
  for (int i = 0; i < N; i++) {
    scanf("%lld", &a[i]);
  }
  solve(N, x, std::move(a));
  return 0;
}
