The cows have once again tried to form a startup company, failing to remember from past experience that cows make terrible managers!
The cows, conveniently numbered 1 \ldots N1…N ( 1 \leq N \leq 100,0001≤N≤100,000 ), organize the company as a tree, with cow 1 as the president (the root of the tree). Each cow except the president has a single manager (its "parent" in the tree). Each cow ii has a distinct proficiency rating, p(i)p(i) , which describes how good she is at her job. If cow ii is an ancestor (e.g., a manager of a manager of a manager) of cow jj , then we say jj is a subordinate of ii .
Unfortunately, the cows find that it is often the case that a manager has less proficiency than several of her subordinates, in which case the manager should consider promoting some of her subordinates. Your task is to help the cows figure out when this is happening. For each cow ii in the company, please count the number of subordinates jj where p(j) > p(i)p(j)>p(i) .
奶牛们又一次试图创建一家创业公司，还是没有从过去的经验中吸取教训--牛是可怕的管理者！
为了方便，把奶牛从 1⋯N(1≤N≤100,000) 编号，把公司组织成一棵树，1 号奶牛作为总裁（这棵树的根节点）。除了总裁以外的每头奶牛都有一个单独的上司（它在树上的 “双亲结点”）。所有的第 ii 头牛都有一个不同的能力指数 p(i)p(i) ，描述了她对其工作的擅长程度。如果奶牛 ii 是奶牛 jj 的祖先节点（例如，上司的上司的上司），那么我们我们把奶牛 jj 叫做 ii 的下属。
不幸地是，奶牛们发现经常发生一个上司比她的一些下属能力低的情况，在这种情况下，上司应当考虑晋升她的一些下属。你的任务是帮助奶牛弄清楚这是什么时候发生的。简而言之，对于公司的中的每一头奶牛 ii ，请计算其下属 jj 的数量满足 p(j) > p(i)p(j)>p(i) 。
输入输出格式输入格式：
The first line of input contains NN .
The next NN lines of input contain the proficiency ratings p(1) \ldots p(N)p(1)…p(N) for the cows. Each is a distinct integer in the range 1 \ldots 1,000,000,0001…1,000,000,000 .