A smile house is created to raise the mood. It has n rooms. Some of the rooms are connected by doors. For each two rooms (number i and j), which are connected by a door, Petya knows their value cij — the value which is being added to his mood when he moves from room i to room j.

Petya wondered whether he can raise his mood infinitely, moving along some cycle? And if he can, then what minimum number of rooms he will need to visit during one period of a cycle?

Input

The first line contains two positive integers n and m (<image>), where n is the number of rooms, and m is the number of doors in the Smile House. Then follows the description of the doors: m lines each containing four integers i, j, cij и cji (1 ≤ i, j ≤ n, i ≠ j, - 104 ≤ cij, cji ≤ 104). It is guaranteed that no more than one door connects any two rooms. No door connects the room with itself.

Output

Print the minimum number of rooms that one needs to visit during one traverse of the cycle that can raise mood infinitely. If such cycle does not exist, print number 0.

Examples

Input

4 4
1 2 -10 3
1 3 1 -10
2 4 -10 -1
3 4 0 -3


Output

4

Note

Cycle is such a sequence of rooms a1, a2, ..., ak, that a1 is connected with a2, a2 is connected with a3, ..., ak - 1 is connected with ak, ak is connected with a1. Some elements of the sequence can coincide, that is, the cycle should not necessarily be simple. The number of rooms in the cycle is considered as k, the sequence's length. Note that the minimum possible length equals two.
5 4
1 2 1 -1
2 3 1 -1
3 1 1 -1
4 5 2 -1
 3 3
1 2 -10 30
1 3 1 1
2 3 -10 -1
 6 15
1 2 -52 -10
1 3 2 -72
1 4 -72 -2
1 5 6 -100
1 6 -100 -97
2 3 -63 -37
2 4 -99 -55
2 5 -84 -9
2 6 -17 -8
3 4 -16 -57
3 5 -1 8
3 6 -22 -88
4 5 -59 -75
4 6 -92 -63
5 6 9 -65
 10 9
1 2 3 -5
2 3 5 -7
3 4 7 -9
4 5 8 -11
5 6 7 -10
1 6 -10 5
2 4 1 -15
4 6 1 -16
2 6 -9 1
 10 8
1 2 3 -5
2 3 5 -7
3 4 7 -9
4 5 8 -11
5 1 7 -10
4 7 -1 -1
7 8 -1 -1
8 4 -1 -1
 2 0
 2 1
1 2 10 10
 7 21
1 2 -91 -94
1 3 4 -96
1 4 -98 -100
1 5 -91 -98
1 6 -97 -29
1 7 -98 -91
2 3 -99 -98
2 4 6 -90
2 5 -96 -95
2 6 -100 -96
2 7 -94 -4
3 4 -95 -97
3 5 11 -94
3 6 -93 -91
3 7 -97 -95
4 5 -91 -93
4 6 1 -98
4 7 -92 -95
5 6 -95 -98
5 7 12 -90
6 7 -97 -99
 300 0
 5 10
1 2 -22 -3
1 3 4 -30
1 4 -30 -30
1 5 -12 -21
2 3 9 -28
2 4 9 -19
2 5 -23 7
3 4 -17 -4
3 5 -4 -21
4 5 -12 -4
 4 6
1 2 -65 -53
1 3 -64 -19
1 4 -77 -64
2 3 -20 8
2 4 -11 -61
3 4 -39 -8
 2 1
1 2 10 -10
 3 3
1 2 -10 3
1 3 1 -10
2 3 -10 -1
 10 9
1 2 3 -5
2 3 5 -7
3 4 7 -9
4 5 8 -11
5 1 7 -10
4 7 -1 -1
7 8 -1 -1
8 4 -1 -1
6 10 1 1
 1 0

2
 2
 2
 3
 5
 0
 2
 7
 0
 3
 0
 0
 3
 2
 0

import java.io.*;
import java.util.*;
import java.math.*;

public class Main implements Runnable {
	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer st;
	private Random rnd;
	
	final String filename = "codeforces";
	
	final int inf = 1000000000;
	final int log = 11;
	
	public void solve() throws IOException {
		int n = nextInt(), m = nextInt();
		
		int[][][] shortPathes = new int[log][n][n];
		int[][] dNotFounded = new int[n][n];
		int[][] dCur = new int[n][n];
		
		for(int i = 0; i < log; i++) for(int j = 0; j < n; j++) Arrays.fill(shortPathes[i][j], inf);
		for(int i = 0; i < n; i++) Arrays.fill(dNotFounded[i], inf);
		for(int i = 0; i < n; i++) dNotFounded[i][i] = shortPathes[0][i][i] = 0;
		
		for(int i = 0; i < m; i++) {
			int u = nextInt() - 1, v = nextInt() - 1;
			
			shortPathes[0][u][v] = -nextInt();
			shortPathes[0][v][u] = -nextInt();
		}
		
		for(int it = 1; it < log; it++) {
			for(int k = 0; k < n; k++) {
				for(int i = 0; i < n; i++) {
					if(shortPathes[it - 1][i][k] == inf) continue;
					
					for(int j = 0; j < n; j++) {
						if(shortPathes[it - 1][k][j] == inf) continue;
						
						int a = shortPathes[it - 1][i][k] + shortPathes[it - 1][k][j];
						
						shortPathes[it][i][j] = (shortPathes[it][i][j] < a) ? shortPathes[it][i][j] : a;
					}
				}
			}
		}
		
		boolean a = false;
		
		for(int i = 0; i < n; i++) if(shortPathes[log - 1][i][i] < 0) a = true;
		
		if(!a) {
			out.println(0);
			return;
		}
		
		int res = 0;
		
		for(int it = log - 1; it >= 0; it--) {
			for(int i = 0; i < n; i++) Arrays.fill(dCur[i], inf);
			boolean finded = false;
			
			search:
			for(int k = 0; k < n; k++) {
				for(int i = 0; i < n; i++) {
					if(shortPathes[it][i][k] == inf) continue;
					
					for(int j = 0; j < n; j++) {
						if(dNotFounded[k][j] == inf) continue;
						
						int b = shortPathes[it][i][k] + dNotFounded[k][j];
						
						dCur[i][j] = (dCur[i][j] < b) ? dCur[i][j] : b; // hint: ����� ���� ��� ����� res + 2^it
					}
					
					if(dCur[i][i] < 0) {
						finded = true;
						break search;
					}
				}
			}
			
			if(!finded) {
				res += (1 << it);
				
				for(int i = 0; i < n; i++) {
					System.arraycopy(dCur[i], 0, dNotFounded[i], 0, n);
				}
			}
		}
		
		// �.�. � ������ ����� res �� ����� �� ������, ��������� ��������
		out.println(res + 1);
	}
		
	public static void main(String[] args) {
		new Main().run();
	} 
	
	public void run() {
		try {
			try {
				in = new BufferedReader(new FileReader(filename + ".in"));
				out = new PrintWriter(new FileWriter(filename + ".out"));
			} catch(FileNotFoundException e) {
				in = new BufferedReader(new InputStreamReader((System.in)));
				out = new PrintWriter(System.out);
			}
			
			st = null;
			rnd = new Random();
			
			solve();
			
			out.close();
		} catch(IOException e) {
			e.printStackTrace();
		}	
	}
	
	private String nextToken() throws IOException, NullPointerException {
		while(st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		
		return st.nextToken();
	}
	
	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	
	private long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	
	private double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

}