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;

vector <int> adia[100010];
int see[100010];
int maxim = 0, total;

int lsb(int x, int i)
{
    for (; ; i++)
        if ((1 << i) & x)
            return i;
}

int msb(int x)
{
    for (int i = 30; ; i--)
        if ((1 << i) & x)
            return i;
}

int dfs(int nod, int tata)
{
    int ce_pun = 0;
    for (auto i : adia[nod]) {
        if (i != tata) {
            int m = dfs(i, nod);
            if (m & see[nod])
                ce_pun = max(ce_pun, msb(m & see[nod]));
            see[nod] |= m;
        }
    }
    ce_pun = lsb(~see[nod], ce_pun);
    maxim = max(maxim, ce_pun);

    see[nod] |= (1 << ce_pun);
    for (int i = 0; i < ce_pun; i++)
        see[nod] &= ~(1 << i);

    return see[nod];
}

int main()
{
    int n, a, b;
    cin >> n;

    while (--n) {
        cin >> a >> b;
        adia[a].push_back(b);
        adia[b].push_back(a);
    }

    dfs(1, 0);

    cout << maxim << '\n';
}
