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 <cstdio>
#include <vector>
static const int MAXN = 1e5 + 4;
static const int LOGN = 18;

int n;
std::vector<int> e[MAXN];

int ans = 0;
int f[MAXN];
void dfs_dp(int u, int p = -1)
{
    int mask = 0, label = 0;
    for (int v : e[u]) if (v != p) {
        dfs_dp(v, u);
        for (int i = label + 1; i < LOGN; ++i)
            if ((f[v] & (1 << i)) && (mask & (1 << i))) label = i;
        mask |= f[v];
    }
    //int label = 8 * sizeof(int) - __builtin_clz(f[u]) - 1;
    while (mask & (1 << label)) ++label;
    f[u] = (1 << label);
    for (int i = label + 1; i < LOGN; ++i)
        if (mask & (1 << i)) f[u] |= (1 << i);
    if (ans < label) ans = label;
}

int main()
{
    scanf("%d", &n);
    int u, v;
    for (int i = 0; i < n - 1; ++i) {
        scanf("%d%d", &u, &v); --u, --v;
        e[u].push_back(v);
        e[v].push_back(u);
    }

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

    return 0;
}
