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 int MOD = 1e9 + 7;
long long POWER[65];
long long power(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    if (ret >= MOD) ret %= MOD;
    if (a >= MOD) a %= MOD;
    b >>= 1;
  }
  return ret;
}
long long inv(long long x) { return power(x, MOD - 2); }
void precompute() {
  POWER[0] = 1;
  for (int i = 1; i < 63; i++) POWER[i] = POWER[i - 1] << 1LL;
}
const int MAXN = 2e3 + 3;
long long ar[MAXN];
int main() {
  long long n, x;
  cin >> n >> x;
  long long mini = 1e18;
  int idx;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &ar[i]);
    if (ar[i] < mini) mini = ar[i], idx = i;
  }
  long long ans = ar[idx];
  int in = idx;
  idx++;
  if (idx == n + 1) idx = 1;
  while (idx != in) {
    if (ar[idx] > ar[in] + x) {
      ans += x;
      ans += ar[in];
    } else
      ans += ar[idx];
    idx++;
    if (idx == n + 1) idx = 1;
  }
  cout << ans << endl;
  return 0;
}
