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 = 100010;
int n, ans, mask[maxn];
vector<int> G[maxn];

int main() {
    scanf("%d", &n);
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d %d", &u, &v);
        G[u].push_back(v), G[v].push_back(u);
    }
    function<void(int, int)> dfs = [&](int v, int fa) {
        vector<int> cnt(20);
        for (int u : G[v]) if (u ^ fa) {
            dfs(u, v), mask[v] |= mask[u];
            for (int i = 0; i < 20; i++) {
                if (mask[u] >> i & 1) cnt[i]++;
            }
        }
        int cur = 0;
        for (int i = 19; ~i; i--) {
            if (cnt[i] > 1) { cur = i + 1; break; }
        }
        while (mask[v] >> cur & 1) cur++;
        mask[v] |= (1 << cur), ans = max(ans, cur);
        mask[v] = mask[v] >> cur << cur;
    };
    dfs(1, 0), printf("%d\n", ans);
    return 0;
}