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;

#define int long long
const double PI = 3.14159265358979323846;
typedef vector<int> vint;
typedef pair<int, int> pint;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
// cout << fixed << setprecision(20);

int N;
int a[110000], b[110000];
vint G[110000];

int dfs(int v, int p) {
    int a = 0;
    vint vec;
    for (int i = 0; i < G[v].size(); i++) {
        if (G[v][i] == p)continue;
        vec.push_back(dfs(G[v][i], v));
        a |= vec[vec.size() - 1];
    }
    int idx;
    for (int i = 20; i >= 0; i--) {
        if ((a & (1 << i)) == 0)idx = i;
        int cnt = 0;
        for (int j = 0; j < vec.size(); j++)cnt += (vec[j] & (1 << i)) ? 1 : 0;
        if (cnt >= 2)break;
    }
    a |= 1 << idx;
    a -= a % (1 << idx);
    return a;
}

signed main() {
    cin >> N;
    for (int i = 0; i < N - 1; i++)cin >> a[i] >> b[i], G[a[i]].push_back(b[i]), G[b[i]].push_back(a[i]);
    int x = dfs(1, 0);
    for (int i = 20; i >= 0; i--) {
        if (x & (1 << i)) {cout << i << endl; break;}
    }
}
