CF137C.History

Polycarpus likes studying at school a lot and he is always diligent about his homework. Polycarpus has never had any problems with natural sciences as his great-great-grandfather was the great physicist Seinstein. On the other hand though, Polycarpus has never had an easy time with history.

Everybody knows that the World history encompasses exactly n events: the i-th event had continued from the year a__i to the year b__i inclusive (a__i < b__i). Polycarpus easily learned the dates when each of n events started and ended (Polycarpus inherited excellent memory from his great-great-granddad). But the teacher gave him a more complicated task: Polycaprus should know when all events began and ended and he should also find out for each event whether it includes another event. Polycarpus' teacher thinks that an event j includes an event i if a__j < a__i and b__i < b__j. Your task is simpler: find the number of events that are included in some other event.

波利卡普斯非常喜欢在学校学习，而且他总是认真完成家庭作业。波利卡普斯在自然科学方面从未遇到过困难，因为他的高祖是一位伟大的物理学家——赛因斯坦。然而，他在历史学科上却一直不太顺利。

众所周知，世界历史恰好包含 $n$ 个事件：其中第 $i$ 个事件从年份 $a_i$ 持续到年份 $b_i$（含端点），即 $a_i < b_i$。波利卡普斯轻松记住了这 $n$ 个事件各自的起止年份（他从高祖那里继承了极佳的记忆力）。但老师给他布置了一项更复杂的任务：波利卡普斯不仅要熟记所有事件的起止时间，还需判断每个事件是否被另一个事件所包含。老师认为：若事件 $j$ 包含事件 $i$，当且仅当 $a_j < a_i$ 且 $b_i < b_j$。而你的任务则更简单：请计算有多少个事件被至少另一个事件所包含。

The first input line contains integer n (1 ≤ n ≤ 105) which represents the number of events. Next n lines contain descriptions of the historical events, one event per line. The i + 1 line contains two integers a__i and b__i (1 ≤ a__i < b__i ≤ 109) — the beginning and the end of the i-th event. No two events start or finish in the same year, that is, a__i ≠ a__j, a__i ≠ b__j, b__i ≠ a__j, b__i ≠ b__j for all i, j (where i ≠ j). Events are given in arbitrary order.

第一行输入包含一个整数 $n$（$1 \leq n \leq 10^5$），表示事件的数量。接下来的 $n$ 行每行描述一个历史事件，第 $i+1$ 行包含两个整数 $a_i$ 和 $b_i$（$1 \leq a_i < b_i \leq 10^9$），分别表示第 $i$ 个事件的起始年份和结束年份。任意两个事件的起始年份或结束年份均不相同，即对所有 $i, j$（其中 $i \neq j$）均有：$a_i \neq a_j$、$a_i \neq b_j$、$b_i \neq a_j$、$b_i \neq b_j$。事件以任意顺序给出。

Print the only integer — the answer to the problem.

输出唯一的整数——该问题的答案。

In the first example the fifth event is contained in the fourth. Similarly, the fourth event is contained in the third, the third — in the second and the second — in the first.

In the second example all events except the first one are contained in the first.

In the third example only one event, so the answer is 0.

在第一个例子中，第五个事件包含在第四个事件中；类似地，第四个事件包含在第三个事件中，第三个事件包含在第二个事件中，第二个事件包含在第一个事件中。

在第二个例子中，除第一个事件外，其余所有事件均包含在第一个事件中。

在第三个例子中，仅有一个事件，因此答案为 0。