Codeforces 1311 F. Moving Points （树状数组+离散化）

Description：

There are $n$ points on a coordinate axis OX. The $i-th$ point is located at the integer point $xi$ and has a speed $vi$ . It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the $i-th$ point at the moment $t$ (t can be non-integer) is calculated as $xi+t⋅vi$ .

Consider two points $i$ and $j$ . Let $d(i,j)$ be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value $d(i,j)$ will be 0.

Your task is to calculate the value $\sum_{1≤i<j≤n}d(i,j)$ (the sum of minimum distances over all pairs of points).

Input

The first line of the input contains one integer $n (2≤n≤2⋅10^5)$ — the number of points.

The second line of the input contains $n$ integers $x_1,x_2,…,x_n (1≤x_i≤10^8)$ , where xi is the initial coordinate of the i-th point. It is guaranteed that all xi are distinct.

The third line of the input contains $n$ integers $v_1,v_2,…,v_n (−10^8≤vi≤10^8)$ , where $v_i$ is the speed of the $i-th$ point.

Output

Print one integer — the value $\sum_{1≤i<j≤n}d(i,j)$ (the sum of minimum distances over all pairs of points).

Examples

input

3
1 3 2
-100 2 3

output

input

5
2 1 4 3 5
2 2 2 3 4

output

input

2
2 1
-3 0

output

解题思路：

给出 $n$ 个点和他们的初始位置 $x$ 和移动速度 $v$ ，问所有点对在任意时刻全部最小值的和是多少，不是同一时刻而是每个点对的最小值的和，可以是很多时刻。

时间可以是任意实数，对于任意两点，如果 $x_i<=x_j$ ,且 $x_i<=v_j$ ，那么 $i，j$ 永远不可能相遇。否则肯定可以相遇，所以只需要计算 $x_i<=x_j$ ,且 $x_i<=v_j$ 的情况即可。这种情况贡献就是距离，树状数组维护一个和就好了。

AC代码：

const int N = 5e3 + 5;
int n, d;
int ans[N];
int ch[N];
int fa[N];
int c, now, mx;
int cnt, pos;
int get_dep()
{
    int ans = now;
    c++;
    if (c == mx)
        now++, mx *= 2, c = 1;
    return ans;
}

int main()
{
    int t;
    sd(t);
    while (t--)
    {
        mem(ch, 0);
        sdd(n, d);
        cnt = 0;
        rep(i, 1, n - 1)
        {
            ans[i] = i;
            cnt += i;
        }
        pos = n - 1;
        now = 1, c = 1, mx = 2;
        if (cnt < d)
        {
            puts("NO");
            continue;
        }
        while (cnt > d)
        {
            int dep = get_dep();
            if (dep >= pos)
                break;
            int tmp = cnt - d;
            if (tmp >= (pos - dep))
            {
                cnt -= pos - dep;
                ans[pos] = dep;
            }
            else
            {
                ans[pos] -= tmp;
                cnt = d;
            }
            pos--;
        }
        if (cnt > d)
            puts("NO");
        else
        {
            puts("YES");
            ans[0] = 0;
            rep(i, 1, n - 1)
            {
                rep(j, 0, n - 1) if (ans[j] == ans[i] - 1 && ch[j] < 2)
                {
                    ch[j]++;
                    fa[i] = j;
                    break;
                }
            }
            rep(i, 1, n - 1)
            {
                printf("%d%c", fa[i] + 1, i == n - 1 ? '\n' : ' ');
            }
        }
    }
    return 0;
}