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 *buildRMQ(int *a, int n) {
  int logn = 1;
  for (int k = 1; k < n; k *= 2) ++logn;
  int *r = new int[n * logn];
  int *b = r;
  copy(a, a + n, b);
  for (int k = 1; k < n; k *= 2) {
    copy(b, b + n, b + n);
    b += n;
    for (int i = (0); i < (n - k); ++i) b[i] = min(b[i], b[i + k]);
  }
  return r;
}
int minimum(int x, int y, int *rmq, int n) {
  int z = y - x, k = 0, e = 1, s;
  s = ((z & 0xffff0000) != 0) << 4;
  z >>= s;
  e <<= s;
  k |= s;
  s = ((z & 0x0000ff00) != 0) << 3;
  z >>= s;
  e <<= s;
  k |= s;
  s = ((z & 0x000000f0) != 0) << 2;
  z >>= s;
  e <<= s;
  k |= s;
  s = ((z & 0x0000000c) != 0) << 1;
  z >>= s;
  e <<= s;
  k |= s;
  s = ((z & 0x00000002) != 0) << 0;
  z >>= s;
  e <<= s;
  k |= s;
  return min(rmq[x + n * k], rmq[y + n * k - e + 1]);
}
int main() {
  int N, x;
  cin >> N >> x;
  int aa[4010];
  int *a = aa + 2000;
  int *b = a - N;
  for (int i = (0); i < (N); ++i) cin >> a[i];
  for (int i = (0); i < (N); ++i) *(a - i - 1) = a[N - 1 - i];
  int *rmq = buildRMQ(b, 2 * N);
  long long ans = (long long)1000000000 * 2000 + 1;
  for (int k = (0); k < (N); ++k) {
    long long cost = 0;
    cost += x * k;
    for (int i = (0); i < (N); ++i) {
      cost += minimum(i + N - k, i + N, rmq, 2 * N);
    }
    ans = ans > cost ? cost : ans;
  }
  cout << ans << endl;
  return 0;
}
