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 long long N = 2e3 + 10, lg = 12;
long long n, x, ar[N], rmq[N][lg], dp[N][N];
long long get(long long a, long long b) {
  long long ans = a;
  for (long long i = 0; i < lg; i++) {
    if ((b >> i) & 1) {
      if (ar[ans] > ar[rmq[a][i]]) ans = rmq[a][i];
      a -= (1 << i);
    }
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> x;
  for (long long i = 0; i < n; i++) cin >> ar[i];
  for (long long i = 1; i <= n - 1; i++) {
    rmq[i][0] = i - 1;
  }
  for (long long j = 1; j < lg; j++)
    for (long long i = 0; i < n; i++) {
      if (i - (1 << j) >= 0) {
        if (ar[rmq[i][j - 1]] > ar[rmq[i - (1 << (j - 1))][j - 1]])
          rmq[i][j] = rmq[i - (1 << (j - 1))][j - 1];
        else
          rmq[i][j] = rmq[i][j - 1];
      }
    }
  for (long long i = 0; i < N; i++)
    fill(dp[i], dp[i] + N, (long long)1e15 + 10);
  for (long long i = 0; i < n; i++) {
    if (i) {
      dp[i][0] = ar[i] + dp[i - 1][0];
    } else
      dp[i][0] = ar[i];
    for (long long j = 1; j <= min(n - 1, i); j++)
      dp[i][j] = min(dp[i - 1][j] + ar[i], dp[i - 1][j] + ar[get(i, j)]);
    for (long long j = i + 1; j <= n - 1; j++) {
      if (i)
        dp[i][j] = min(dp[i - 1][j] + ar[i],
                       dp[i - 1][j] + ar[get(n - 1, j - (i + 1))]);
      else
        dp[i][j] = min(ar[0], ar[get(n - 1, j - (i + 1))]);
    }
  }
  long long ans = (long long)1e15 + 10;
  for (long long i = 0; i < n; i++) ans = min(ans, dp[n - 1][i] + i * x);
  cout << ans;
}
