You're given a tree with *n* vertices rooted at 1.

We say that there's a *k*-ary heap of depth *m* located at *u* if the following holds:
 -  For *m* = 1 *u* itself is a *k*-ary heap of depth 1. -  For *m* &gt; 1 vertex *u* is a *k*-ary heap of depth *m* if at least *k* of its children are *k*-ary heaps of depth at least *m* - 1. 
Denote *dp*<sub class="lower-index">*k*</sub>(*u*) as maximum depth of *k*-ary heap in the subtree of *u* (including *u*). Your goal is to compute <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/021432cac1fbfdf91e34779ff99d535e19729d02.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The first line contains an integer *n* denoting the size of the tree (2 ≤ *n* ≤ 3·10<sup class="upper-index">5</sup>). 

The next *n* - 1 lines contain two integers *u*, *v* each, describing vertices connected by *i*-th edge.

It's guaranteed that the given configuration forms a tree.
Consider sample case one.

For *k* ≥ 3 all *dp*<sub class="lower-index">*k*</sub> will be equal to 1.

For *k* = 2 *dp*<sub class="lower-index">*k*</sub> is 2 if <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a527e618740bc1e83327ce591b684aa4351c914.png" style="max-width: 100.0%;max-height: 100.0%;"/> and 1 otherwise.

For *k* = 1 *dp*<sub class="lower-index">*k*</sub> values are (3, 1, 2, 1) respectively.

To sum up, 4·1 + 4·1 + 2·2 + 2·1 + 3 + 1 + 2 + 1 = 21.
print("_RANDOM_GUESS_1692698978.591677")# 1692698978.5916927