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 <algorithm>
#include <iostream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <iomanip>
#define _USE_MATH_DEFINES
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
double pi = 3.141592653589793;
ll mod = 1000000007;
int intmax = 2147483647;
int intmin = -2147483648;
ll llmax = 9223372036854775807;
ll llmin = -9223372036854775808;

ll m[2001][2001];

ll modminus(ll x, ll y) {
    return (x - y + mod) % mod;
}

int main() {
    ll N, x;
    cin >> N >> x;
    mod = N;
    for (int i = 0; i < N; i++) {
        cin >> m[0][i];
    }
    for (int n = 1; n < N; n++) {
        for (int i = 0; i < N; i++) {
            m[n][i] = min(m[n - 1][i], m[n - 1][modminus(i , 1)]);
        }
    }
    ll ans = llmax;
    for (int n = 0; n < N; n++) {
        ll num = accumulate(m[n], m[n] + N, 0) + x * n;
        ans = min(ans, num);
    }
    cout << ans << endl;
}
