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;
const int maxN = 1e5 + 13;
int n, ans, val[maxN][50 + 13];
vector<int> adj[maxN];
void dfs (int x, int par) {
    for (auto a : adj[x]) {
	if (a != par) {
	    dfs(a, x);
	    for (int i = 0; i <= 50; i++)
		val[x][i] += val[a][i];
	}
    }
    int cur = 0;
    for (int i = 0; i <= 50; i++) {
	//cout << x << " " << i << " " << val[x][i] << endl;
	if (val[x][i] >= 2)
	    cur = i + 1;
    }
    while (val[x][cur])
	cur++;
    ans = max(ans, cur);
    for (int i = 0; i < cur; i++)
	val[x][i] = 0;
    //cout << x << " " << cur << endl;
    val[x][cur] = 1;
}

int main () {
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
	int a, b;
	cin >> a >> b;
	adj[a].push_back(b);
	adj[b].push_back(a);
    }

    dfs(1, 1);
    cout << ans << endl;
}
