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 "iostream"
#include "algorithm"
#include "string"
#include "vector"
#include "cmath"
#include "bitset"
#include "queue"
#include "functional"
#include "map"
#include "unordered_map"

#define lp(n) for (int i = 0; i < n; i++)
#define LP(n,i) for (int i = 0; i < n; i++)
#define mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nya_n "(=^･ω･^=)"
typedef long long ll;
using namespace std;


ll n, x, a[2000], mem, mx2[2000];
pair<ll, int>mx[2000];
ll cnt;

int main() {
	cin >> n >> x;
	lp(n) {
		cin >> a[i];
		mx[i] = make_pair(a[i], 0);
	}
	lp(n) {
		LP(n, j) {
			if (mx[j].first + x <= mx[(j + 1) % n].first) {
				mx[(j + 1) % n] = make_pair(mx[j].first + x, mx[j].second + 1);
			}
		}
	}
	mem = -1;
	lp(n) {
		mem = max(mem, mx[i].second);
		mx2[i] = a[i];
	}
	lp(n) {
		LP(mem,j) {
			mx2[(i + mem) % n] = min(mx2[(i + mem) % n], a[(i + j) % n]);
		}
	}
	cnt = ll(mem)*ll(x);
	lp(n) {
//		cout << i << sp << mx2[i] << endl;;
		cnt += mx2[i];
	}
	cout << cnt << endl;
	return 0;
}