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
//
// Created by yjq on 2019/11/5.
//

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ld long double
#define ull unsigned long long
#define __ ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

#define  int ll
const int maxn = 4010;
int a[maxn * 2], n, x, r[maxn];
ll s[maxn * 2];
bool vis[maxn * 2];

int c[maxn * 2];

inline int lb(int x) {
    return x & -x;
}

int read(int x) {
    int tot = 0;
    for (; x; x -= lb(x)) {
        tot += c[x];
    }
    return tot;
}

void add(int num, int x) {
    for (; x <= n + n; x += lb(x))c[x] += num;
}

const int cmp(const int i, const int j) {
    return a[i] < a[j];
}

bool can(int x, int q) {
    int left, right;
    int l = 1, r = x;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (read(x) - read(mid - 1) < x - mid + 1)l = mid + 1;
        else r = mid;
    }
    left = l;
    l = x, r = n + n;
    while (l < r) {
        int mid = (l + r + 1) >> 1;
        if (read(mid) - read(x - 1) < mid - x + 1)r = mid - 1;
        else l = mid;
    }
    right = l;
    return right - left + 1 <= q;
}

ll get_ans(int t) {
    for (int i = 1; i <= n; ++i)r[i] = i;
    memset(vis, 0, sizeof vis);
    memset(c, 0, sizeof c);
    sort(r + 1, r + 1 + n, cmp);
    for (int i = n; i >= 1; --i) {
        add(1, r[i]);
        add(1, n + r[i]);
        if (can(r[i], t) && can(n + r[i], t)) {
            vis[r[i]] = vis[n + r[i]] = 1;
        } else {
            add(-1, r[i]);
            add(-1, n + r[i]);
        }
    }
    ll ans = s[n] + 1ll * t * x;
    int b = 2;
    bool ok = 0;
    for (; b <= n + n; ++b) {
        if (vis[b] == 0 && vis[b - 1] == 1) {
            ok = 1;
            break;

        }
    }
    int p = b;
    if (!ok)p = 1;
    for (int i = b; i <= b + n - 1; ++i) {
        if (vis[i])ans -= a[i] - a[p];
        if (vis[i] == 0 && vis[i - 1])p = b;
    }
    return ans;
}

//int
main() {
    __;
    cin >> n >> x;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }
    for (int i = 1; i <= n + n; ++i) {
        if (i > n)a[i] = a[i - n];
        s[i] = s[i - 1] + a[i];
    }
    ll ans = s[n];
    for (int y = 0; y < n; ++y) {
        ans = min(ans, get_ans(y));
    }
    cout << ans << endl;
    return 0;
}