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;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
using ll = long long;
using P = pair<ll, ll>;
using graph = vector<vector<int>>;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ll INF = 1LL << 60;
const ll mod = 998244353LL;
int main() {
  ll N, x;
  cin >> N >> x;
  vector<ll> v(N);
  for (int i = 0; i < (int)(N); i++) cin >> v[i];
  vector<ll> from(N), cost(N);
  for (int i = 0; i < (int)(N); i++) cost[i] = v[i];
  for (int i = 0; i < (int)(N); i++) {
    ll cnt = 0;
    while (cost[(i + cnt) % N] >= v[i] + cnt * x) {
      cost[(i + cnt) % N] = v[i] + cnt * x;
      from[(i + cnt) % N] = i;
      cnt++;
    }
  }
  vector<ll> to(N, 0LL);
  for (int i = 0; i < (int)(N); i++) {
    to[from[i]]++;
  }
  ll ans = 0;
  for (int i = 0; i < (int)(N); i++) {
    if (to[i] != 0) ans += v[i] * to[i] + x * (to[i] - 1LL);
  }
  cout << ans << endl;
}
