Problem: Consider a string with rings on both ends. Rings have positive integers to distinguish them. Rings on both ends of the string have different numbers a and b. Describe this as [a, b]. If there are multiple strings and the number attached to the ring of one string and the other string is the same, these strings can be connected at that ring, and the one made by connecting is called a chain. For example, the strings [1, 3] and [3, 4] form the chain [1, 3, 4]. The string and the chain or the chains are also connected at the ring with the same integer. Suppose you can.

For example, a chain [1, 3, 4] and a string [5, 1] ​​can form [5, 1, 3, 4], and a chain [1, 3, 4] and a chain [2, 3, 5] can form [5, 1, 3, 4]. The chain [1, 3, 4] and the chain [4, 6, 1] form a ring.

There are various shapes like this, but in some of them, rings with the same number follow only once. Multiple connected strings are called chains. For example, chains [1, 3, 4] The centrally crossed shape of the chains [2, 3, 5] also includes the chains [1, 3, 5], [2, 3, 4], and the chains [1, 3, 4]. And chains [4, 6, 1] loops include chains such as [1, 3, 4, 6], [3, 4, 6, 1], [4, 6, 1, 3]. ..

For these chains, the number contained is defined as the length. For a given number of strings, anything that can be connected can form a shape that contains one or more chains. Create a program to find the length of the chain.

The first line of the input data contains the number of strings, the positive integer 1 ≤ n ≤ 100, and each of the following n lines contains two integers a, b separated by blanks 1 ≤ a <b ≤ 100. The two integers in each line represent the integers at both ends of one string.

Input example 1 | Input example 2 | Input example 3
--- | --- | ---
|
7 | 6 | 7
1 3 | 1 2 | 1 3
3 4 | 2 3 | 2 4
1 4 | 3 4 | 3 5
2 7 | 4 5 | 4 6
5 7 | 1 5 | 6 7
6 7 | 2 6 | 2 6
1 7 | | 4 7
Output example 1 | Output example 2 | Output example 3
5 | 6 | 4

input

The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 10.

output

For each dataset, the maximum chain length is output on one line.





Example

Input

7
1 3
3 4
1 4
2 7
5 7
6 7
1 7
6
1 2
2 3
3 4
4 5
1 5
2 6
7
1 3
2 4
3 5
4 6
6 7
2 6
4 7
0


Output

5
6
4

Solution (in CPP):

Solution:
// tsukasa_diary's programing contest code template
#include <bits/stdc++.h>
using namespace std;
// define
#define for_(i,a,b) for(int i=a;i<b;++i)
#define for_rev(i,a,b) for(int i=a;i>=b;--i)
#define allof(a) a.begin(),a.end()
#define minit(a,b) memset(a,b,sizeof(a))
#define size_of(a) (int)a.size()
// typedef
typedef long long lint;
typedef double Double;
typedef pair<int, int> pii;
//
typedef vector< int > Array;
typedef vector< Array > Matrix;
typedef vector< Double > DArray;
typedef vector< DArray > DMatrix;
typedef vector< string > SArray;
typedef vector< pii > PArray;
// popcount
inline int POPCNT(int _x) { return __builtin_popcount(_x); }
inline int POPCNT(lint _x) { return __builtin_popcountll(_x); }
// inf
const int iINF = 1L << 30;
const lint lINF = 1LL << 60;
// eps
Double EPS = 1e-9;
// in range
inline bool in_range(int _v, int _mx, int _mi) { return _mi <= _v && _v < _mx; }
inline bool in_range(Double _v, Double _mi, Double _mx) { return -EPS < _v - _mi && _v - _mx < EPS; }
inline bool in_range(int _x, int _y, int _W, int _H) { return 0 <= _x && _x < _W && 0 <= _y && _y < _H; }
// neighbor clockwise
const int DX[4] = {0,1,0,-1}, DY[4] = {-1,0,1,0};
const int DX_[8] = {0,1,1,1,0,-1,-1,-1}, DY_[8] = {-1,-1,0,1,1,1,0,-1};
// variable update
inline void modAdd(int& _a, int _b, int _m) { _a = (_a + _b) % _m; }
inline void modAdd(lint& _a, lint _b, lint _m) { _a = (_a + _b) % _m; }
inline void minUpdate(int& _a, int _b) { _a = min(_a, _b); }
inline void minUpdate(lint& _a, lint _b) { _a = min(_a, _b); }
inline void maxUpdate(int& _a, int _b) { _a = max(_a, _b); }
inline void maxUpdate(lint& _a, lint _b) { _a = max(_a, _b); }

int n, a[110], b[110];

vector< Array > edges;
bool visit[110];

int rec(int v) {
	int res = 1;
	
	visit[v] = 1;
	
	int e_size = size_of(edges[v]);
	for_(i,0,e_size) {
		int u = edges[v][i];
		if (visit[u]) continue;
		maxUpdate(res, rec(u) + 1);
	}
	
	visit[v] = 0;
	
	return res;
}

void solve() {
	edges.assign(101, Array());
	
	for_(i,0,n) {
		edges[ a[i] ].push_back(b[i]);
		edges[ b[i] ].push_back(a[i]);
	}
	
	minit(visit, 0);
	int ans = 0;
	for_(i,1,101) maxUpdate(ans, rec(i));
	cout << ans << endl;
}

int main() {
	while (cin >> n, n) {
		for_(i,0,n) cin >> a[i] >> b[i];
		solve();
	}
	return 0;
}