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 FILE_IO

int N, ans;
vector <int> edg[100005];
int bts[100005];

void DFS(int nod, int fth)
{
    int msk = 0;
    int val = 0;
    for(auto nxt: edg[nod])
    {
        if(nxt == fth)  continue;
        DFS(nxt, nod);
        for(int b = val + 1; b < 20; b++)
            if( ((1 << b) & msk) && ((1 << b) & bts[nxt]) )
                val = b + 1;
        msk |= bts[nxt];
    }

    while( (1 << val) & msk )   val++;
    bts[nod] = 1 << val;
    for(int b = val + 1; b < 20; b++)
        if( (1 << b) & msk )
            bts[nod] |= (1 << b);
    ans = max(ans, val);
}

int main()
{
    #ifdef FILE_IO
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    #endif

    scanf("%d", &N);
    for(int i = 1; i < N; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        edg[x].push_back(y);
        edg[y].push_back(x);
    }

    DFS(1, 0);
    printf("%d\n", ans);

    return 0;
}
