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;
template <typename T>
void chmax(T &a, T b) {
  a = max(a, b);
}
template <typename T>
void chmin(T &a, T b) {
  a = min(a, b);
}
template <typename T>
void chadd(T &a, T b) {
  a = a + b;
}
long long int dx[] = {0, 0, 1, -1};
long long int dy[] = {1, -1, 0, 0};
constexpr long long INF = 1001001001001001LL;
constexpr long long MOD = 1000000007LL;
long long int N, x;
long long int a[2010];
long long int dp[2010][2010];
signed main() {
  cin >> N >> x;
  for (long long int(i) = (0); (i) < (2010); (i)++)
    for (long long int(j) = (0); (j) < (2010); (j)++) dp[i][j] = INF;
  for (long long int(i) = (0); (i) < (N); (i)++) {
    cin >> a[i];
    dp[i][0] = a[i];
  }
  for (long long int(j) = (1); (j) < (2010); (j)++)
    for (long long int(i) = (0); (i) < (N); (i)++) {
      dp[i][j] = dp[(i - 1 + N) % N][j - 1] + x;
    }
  long long int ans = 0, ma = 0;
  for (long long int(i) = (0); (i) < (N); (i)++) {
    long long int temp = INF, rec = 0, magic = 0;
    for (long long int(j) = (0); (j) < (2010); (j)++) {
      if (temp > dp[i][j]) {
        temp = dp[i][j];
        rec = dp[i][j] - j * x;
        magic = j;
      }
    }
    chmax(ma, magic);
    ans += rec;
  }
  ans += x * ma;
  cout << ans << endl;
  return 0;
}
