地铁系统跟踪不同车站之间的乘客出行时间，并使用这一数据来计算从一站到另一站的平均时间。

实现 UndergroundSystem 类：

• void checkIn(int id, string stationName, int t)
  • 通行卡 ID 等于 id 的乘客，在时间 t ，从 stationName 站进入
  • 乘客一次只能从一个站进入
• void checkOut(int id, string stationName, int t)
  • 通行卡 ID 等于 id 的乘客，在时间 t ，从 stationName 站离开
• double getAverageTime(string startStation, string endStation)
  • 返回从 startStation 站到 endStation 站的平均时间
  • 平均时间会根据截至目前所有从 startStation 站 直接 到达 endStation 站的行程进行计算，也就是从 startStation 站进入并从 endStation 离开的行程
  • 从 startStation 到 endStation 的行程时间与从 endStation 到 startStation 的行程时间可能不同
  • 在调用 getAverageTime 之前，至少有一名乘客从 startStation 站到达 endStation 站

你可以假设对 checkIn 和 checkOut 方法的所有调用都是符合逻辑的。如果一名乘客在时间 t1 进站、时间 t2 出站，那么 t1 < t2 。所有时间都按时间顺序发生。

示例 1：

输入
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]

输出
[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]

解释
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15);  // 乘客 45 "Leyton" -> "Waterloo" ，用时 15-3 = 12
undergroundSystem.checkOut(27, "Waterloo", 20);  // 乘客 27 "Leyton" -> "Waterloo" ，用时 20-10 = 10
undergroundSystem.checkOut(32, "Cambridge", 22); // 乘客 32 "Paradise" -> "Cambridge" ，用时 22-8 = 14
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // 返回 14.00000 。只有一个 "Paradise" -> "Cambridge" 的行程，(14) / 1 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // 返回 11.00000 。有两个 "Leyton" -> "Waterloo" 的行程，(10 + 12) / 2 = 11
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // 返回 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38);  // 乘客 10 "Leyton" -> "Waterloo" ，用时 38-24 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // 返回 12.00000 。有三个 "Leyton" -> "Waterloo" 的行程，(10 + 12 + 14) / 3 = 12

示例 2：

输入
["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]

输出
[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]

解释
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(10, "Leyton", 3);
undergroundSystem.checkOut(10, "Paradise", 8); // 乘客 10 "Leyton" -> "Paradise" ，用时 8-3 = 5
undergroundSystem.getAverageTime("Leyton", "Paradise"); // 返回 5.00000 ，(5) / 1 = 5
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16); // 乘客 5 "Leyton" -> "Paradise" ，用时 16-10 = 6
undergroundSystem.getAverageTime("Leyton", "Paradise"); // 返回 5.50000 ，(5 + 6) / 2 = 5.5
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30); // 乘客 2 "Leyton" -> "Paradise" ，用时 30-21 = 9
undergroundSystem.getAverageTime("Leyton", "Paradise"); // 返回 6.66667 ，(5 + 6 + 9) / 3 = 6.66667

提示：

• 1 <= id, t <= 106
• 1 <= stationName.length, startStation.length, endStation.length <= 10 次
• 所有字符串由大小写英文字母与数字组成
• 总共最多调用 checkIn、checkOut 和 getAverageTime 方法 2 * 104
• 与标准答案误差在 10-5 以内的结果都被视为正确结果