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 <cstdio>
#include <algorithm>
#include <cctype>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5;
struct Edge
{
    int next, to;
    Edge(int next = 0, int to = 0):next(next), to(to) {};
}edge[N<<1];
int tot, head[N];
void _add(int x, int y) { edge[++tot] = Edge(head[x], y), head[x] = tot; }
void add(int x, int y) { _add(x, y), _add(y, x); }
int n, f[N], ans;
void dfs(int x, int fa)
{
    int s = 0, d = 20;
    for(int i = head[x]; i; i = edge[i].next)
    {
        int y = edge[i].to; if(y == fa) continue;
        dfs(y, x), s |= f[x]&f[y], f[x] |= f[y];
    }
    while(~d&&~s>>d&1) --d; ++d; while(f[x]>>d&1) ++d;
    f[x] = (f[x]>>d|1)<<d, ans = max(ans, d);
}
int main()
{
    n = read();
    for(int i = 1, x, y; i < n; ++i) x = read(), y = read(), add(x, y);
    dfs(1, 0), printf("%d\n", ans);
    return 0;
}