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>
inline int modexp(int a, int b);
using namespace std;
const int MOD(1e9 + 7);
int n, x, a[4016];
long long d[4016][4016];
long long Poke(int l, int r) {
  if (r - l + 1 >= n) return 0;
  long long &ret = d[l][r];
  if (~ret) return ret;
  ret = 1LL << 59;
  if (l > 1) ret = min(ret, Poke(l - 1, r) + a[l - 1]);
  if (r < n + n) ret = min(ret, Poke(l, r + 1) + a[r + 1]);
  if (r < n + n) ret = min(ret, Poke(l + 1, r + 1) + x);
  return ret;
}
int main() {
  scanf("%d%d", &n, &x);
  memset(d, -1, sizeof d);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) a[n + i] = a[i];
  long long ans = 1LL << 59;
  for (int i = 1; i <= n; i++) ans = min(ans, Poke(i, i) + a[i]);
  printf("%lld\n", ans);
  return 0;
}
inline int modexp(int a, int b) {
  int e = a, r = 1;
  while (b > 0) {
    if (b & 1) r = r * 1LL * e % MOD;
    e = e * 1LL * e % MOD;
    b >>= 1;
  }
  return r;
}
