We will recursively define uninity of a tree, as follows: (Uni is a Japanese word for sea urchins.)

* A tree consisting of one vertex is a tree of uninity 0.
* Suppose there are zero or more trees of uninity k, and a vertex v. If a vertex is selected from each tree of uninity k and connected to v with an edge, the resulting tree is a tree of uninity k+1.



It can be shown that a tree of uninity k is also a tree of uninity k+1,k+2,..., and so forth.

You are given a tree consisting of N vertices. The vertices of the tree are numbered 1 through N, and the i-th of the N-1 edges connects vertices a_i and b_i.

Find the minimum k such that the given tree is a tree of uninity k.

Constraints

* 2 ≦ N ≦ 10^5
* 1 ≦ a_i, b_i ≦ N(1 ≦ i ≦ N-1)
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

Print the minimum k such that the given tree is a tree of uninity k.

Examples

Input

7
1 2
2 3
2 4
4 6
6 7
7 5


Output

2


Input

12
1 2
2 3
2 4
4 5
5 6
6 7
7 8
5 9
9 10
10 11
11 12


Output

3
#include<bits/stdc++.h>

using namespace std;

int N, dp[100009];
char col[100009], lg[400009], sml[400009];
vector < int > v[100009];

void dfs (int nod, int tata)
{
    int sau = 0;
    for (auto son : v[nod])
        if (son != tata)
        {
            dfs (son, nod);
            int curr = dp[son] & sau;
            if (curr > 0)
                sau |= ((1 << (lg[curr] + 1)) - 1);
            sau |= dp[son];
        }
    col[nod] = sml[sau];
    dp[nod] = ((sau >> col[nod]) << col[nod]) | (1 << col[nod]);
}


int main ()
{
scanf ("%d", &N);
for (int i=1; i<N; i++)
{
    int x, y;
    scanf ("%d %d", &x, &y);
    v[x].push_back (y);
    v[y].push_back (x);
}
for (int i=2; i<=4 * N; i++)
    lg[i] = lg[i >> 1] + 1;
for (int i=1; i<=4 * N; i++)
    if (i & 1) sml[i] = sml[i >> 1] + 1;
    else sml[i] = 0;
dfs (1, -1);
printf ("%d\n", lg[dp[1]]);
return 0;
}
