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>
const long long maxn = 2e3 + 10;
inline long long Max(long long a, long long b) { return a > b ? a : b; }
inline long long Min(long long a, long long b) { return a > b ? b : a; }
namespace FastIO {
inline long long read() {
  long long f = 1, r = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') {
      f = -1;
    }
    c = getchar();
  }
  while (isdigit(c)) {
    r = 10 * r + c - '0';
    c = getchar();
  }
  return f * r;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) { write(x), puts(""); }
};  // namespace FastIO
using namespace FastIO;
long long N, X, arr[maxn];
struct Node {
  long long base, transe;
} nodes[maxn];
signed main() {
  N = read(), X = read();
  for (register long long i = 1; i <= N; ++i) {
    arr[i] = read();
  }
  long long mxt = 0;
  for (register long long i = 1; i <= N; ++i) {
    long long mx = 0x3f3f3f3f3f3f3f3f, id = i, tr = 0;
    for (register long long j = 1; j <= N; ++j) {
      long long t = j > i ? (N - (j - i)) : (i - j), cost = t * X + arr[j];
      if (cost < mx || (cost == mx && t < tr)) {
        mx = cost, id = j, tr = t;
      }
    }
    nodes[i].base = arr[id], nodes[i].transe = tr, mxt = Max(mxt, tr);
  }
  long long sum = 0;
  for (register long long i = 1; i <= N; ++i) {
    sum += nodes[i].base;
  }
  sum += mxt * X;
  writeln(sum);
  return 0;
}
