K. Wolf and sheep

A sheep lives on an infinite field. The sheep wants to eat some grass. Grass only exists in one place. That place is a circle defined by a center point (G_x, G_y) and a radius G_r.

The sheep would gladly eat all the grass. But if you read the title of this task you noticed that there is also a wolf on that field. The wolf wants to eat the sheep, but we don't want that to happen.

The wolf is tied to a pole at position (W_x, W_y) with a rope of length W_r.

Find the area of grass that the sheep can eat without being eaten by the wolf. We can assume that the sheep starts in a safe location.

Input

The first line contains the number of test cases T (1 ≤ T ≤ 10000).

Each test case consists of 2 lines. The first line contains integers G_x, G_y and G_r. The second line contains integers W_x, W_y and W_r. ( - 10⁵ ≤ G_x, G_y, W_x, W_y ≤ 10⁵, 1 ≤ G_r, W_r ≤ 10⁵)

Output

For each test case output one line containing “Case #tc: A”, where tc is the number of the test case (starting from 1) and A is the area of grass that the sheep can eat safely with a maximum absolute or relative error of 10^- 6.

Examples

Input

2
0 0 5
0 10 5
0 0 5
5 0 5

Output

Case #1: 78.53981634
Case #2: 47.83057387

有一头羊和一头狼，分别有一个园的范围，在狼会吃羊，求羊最大可以吃草的范围是多大。

1.当两园相离或外切时，答案就是羊所在的面积

2.当狼的范围包含羊时，答案就是0

3.当羊的范围包含狼是，答案就是羊的面积减去狼的面积

4.当两园相交时，答案就是羊的面积减去重合的面积

主要是相交的面积难算，当相交时，重合面积是两个扇形面积之和减去平行四边形的面积。可以用余弦定理算出两个扇形的弧度，这个扇形面积和平行四边形的面积就出来了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <vector>
#include <set>
#include <map>
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int N = 110;
double dis(double x1, double y1, double x2, double y2) {
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}
int main() {
    int t, k = 1;
    scanf("%d", &t);
    double pi = acos(-1);
    while(t--) {
        double gx, gy, gr, wx, wy, wr;
        cin >> gx >> gy >> gr;
        cin >> wx >> wy >> wr;
        double dd = dis(gx, gy, wx, wy);
        if(dd >= (gr+wr)) {
            printf("Case #%d: %.8lf\n",k++,pi*gr*gr);
        } else if(dd + wr <= gr ){
            printf("Case #%d: %.8lf\n",k++,pi*(gr*gr - wr*wr));
        } else if(dd + gr <= wr) {
            printf("Case #%d: 0.00000000\n",k++);
        } else {
            double a1 = acos((gr*gr + dd*dd - wr*wr) / (2*gr*dd));
            double a2 = acos((wr*wr + dd*dd - gr*gr) / (2*wr*dd));
            double area1 = (sin(a1*2)*gr*gr+sin(a2*2)*wr*wr)/2;
            double area2 = gr*gr*a1 + wr*wr*a2;
            printf("Case #%d: %.8lf\n",k++,pi*gr*gr-(area2-area1));
        }
    }
    return 0;
}