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 maxn = 2005 + 5;
const double PI = acos(-1.0);
const long long mod = 1e6 + 3;
long long po(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  } else {
    return gcd(b % a, a);
  }
}
long long n, x;
struct no {
  long long id;
  long long val;
} node[maxn];
long long val[maxn], vis[maxn];
bool cmp(no a, no b) { return a.val < b.val; }
void solve() {
  scanf("%lld%lld", &n, &x);
  long long minn = 0x3f3f3f3f;
  long long ans = 0;
  int num = 0;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &val[i]);
    if (val[i] < minn) {
      minn = min(minn, val[i]);
      k = i;
    }
    if (val[i] <= x) {
      num++;
      vis[i] = 1;
      ans += val[i];
    }
  }
  if (num == 0) {
    ans += minn;
    vis[k] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (val[i] <= x + minn) {
      num++;
      vis[i] = 1;
      ans += val[i];
    }
  }
  while (true) {
    int f = 0;
    int k2 = 0;
    long long minn2 = 0x3f3f3f3f;
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        minn2 = min(minn2, val[i]);
        k2 = i;
        f = 1;
      }
    }
    if (f == 0) {
      break;
    }
    if (minn2 <= minn + x) {
      ans += minn2;
      vis[k2] = 1;
      continue;
    }
    int tmp2 = vis[n];
    for (int i = n; i >= 2; i--) {
      vis[i] = vis[i - 1];
    }
    vis[1] = tmp2;
    ans += x;
    minn = 0x3f3f3f3f;
    int k = 0;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      if (minn > val[i]) {
        k = i;
        minn = min(minn, val[i]);
      }
    }
    vis[k] = 1;
    ans += minn;
  }
  printf("%lld", ans);
}
int main() {
  solve();
  return 0;
}
