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 read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!(c >= '0' && c <= '9')) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ((c >= '0' && c <= '9'))
    x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
const int N = 2005;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int n, x;
int a[N];
int sk[N << 1], pos[N], tp, tl;
inline int P(int x, int y) {
  if (x <= y) return y - x;
  return y - x + n;
}
int main() {
  n = read();
  x = read();
  for (register int i = 1; i <= n; i++) a[i] = read();
  long long ans = inf;
  for (register int i = 0; i < n; i++) {
    tl = tp = 0;
    long long tmp = 0;
    for (register int j = i; j >= 1; j--) {
      sk[tp] = a[n - j + 1], pos[tp++] = n - j + 1;
      while (tl < tp - 1 && sk[tp - 1] > sk[tp - 2]) tp--;
    }
    for (register int j = 1; j <= n; j++) {
      sk[tp] = a[j], pos[tp++] = j;
      while (tl < tp && P(pos[tl], j) > i) tl++;
      while (tl < tp - 1 && sk[tp - 1] > sk[tp - 2]) tp--;
      tmp += sk[tp - 1];
    }
    ans = min(ans, tmp + 1ll * i * x);
  }
  printf("%lld\n", ans);
  return 0;
}
