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 <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long

#define maxN 100011
#define inf (17 * maxN)


const int mask = (1 << 30) - 1;

int n, i, x, y;
vector<int> list[maxN];
int conf[maxN];

void solve(int node, int root) {
    int mand, mor, pos;

    mand = 0;
    mor = 0;

    for (auto to : list[node]) {
        if (to == root) continue;
        solve(to, node);

        mand |= (mor & conf[to]);
        mor |= conf[to];
    }


    for (pos = 0; (1 << pos) <= mand; pos++);
    for (; (1 << pos) & mor; pos++);

    mor |= 1 << pos;
    for (pos--; pos >= 0; pos--) mor &= mask ^ (1 << pos);
    conf[node] = mor;
}

int main()
{
    //freopen("test.in","r",stdin);

    scanf("%d", &n);
    for (i = 1; i < n; i++) {
        scanf("%d%d", &x, &y);
        list[x].pb(y);
        list[y].pb(x);
    }

    solve(1, 0);

    int ans = 0;
    for (; (1 << ans) <= conf[1]; ans++);

    printf("%d", ans - 1);


    return 0;
}
