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>
#define rep(i,n) for(int i=0;i<n;i++)
//#define int long long
typedef long long ll;
using namespace std;

ll N, x;
ll a[2000];
ll magic[2001];
pair<ll, ll> d[2010];

signed main()
{
	cin >> N >> x;
	rep(i,N) {
		cin >> a[i];
		d[i].first = a[i];	//コスト
		d[i].second = 0;	//魔法の回数
	}

	/*
	rep(i,N) {
		printf("%d cost = %d, magic = %d\n", i + 1, d[i].first, d[i].second);
	}
	*/

	bool change = true;
	while (change) {
		change = false;
		rep(i,N) {
			if (i == 0) {
				if (d[i].first > d[N-1].first + x) {
					d[i].first = d[N-1].first + x;
					d[i].second += d[N-1].second + 1;
					change = true;
				}
			} else {
				if (d[i].first > d[i-1].first + x) {
					d[i].first = d[i-1].first + x;
					d[i].second += d[i-1].second + 1;
					change = true;
				}
			}
		}
	}

	/*
	rep(i,N) {
		printf("%d cost = %d, magic = %d\n", i + 1, d[i].first, d[i].second);
	}
	*/
	

	rep(i,2001) magic[i] = 0;

	ll ans = 0;
	int magic_max_cnt = 0;
	rep(i,N) {
		ans += d[i].first - x * d[i].second;
		magic_max_cnt = max(magic_max_cnt, d[i].second);
		magic[d[i].second]++;
	}
	/*
	printf("magic_max_cnt = %d\n", magic_max_cnt);
	rep(i,10) {
		printf("magic[%d] = %d\n", i, magic[i]);
	}*/

	/*
	for (int i=1; i<=magic_max_cnt; i++) {
		ans -= x * magic[i];
	}
	*/
	ans += x * magic_max_cnt;

	cout << ans << endl;
}
