Problem: Imagine that there is a group of three friends: A, B and С. A owes B 20 rubles and B owes C 20 rubles. The total sum of the debts is 40 rubles. You can see that the debts are not organized in a very optimal manner. Let's rearrange them like that: assume that A owes C 20 rubles and B doesn't owe anything to anybody. The debts still mean the same but the total sum of the debts now equals 20 rubles.

This task is a generalisation of a described example. Imagine that your group of friends has n people and you know the debts between the people. Optimize the given debts without changing their meaning. In other words, finally for each friend the difference between the total money he should give and the total money he should take must be the same. Print the minimum sum of all debts in the optimal rearrangement of the debts. See the notes to the test samples to better understand the problem.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100; 0 ≤ m ≤ 104). The next m lines contain the debts. The i-th line contains three integers ai, bi, ci (1 ≤ ai, bi ≤ n; ai ≠ bi; 1 ≤ ci ≤ 100), which mean that person ai owes person bi ci rubles.

Assume that the people are numbered by integers from 1 to n.

It is guaranteed that the same pair of people occurs at most once in the input. The input doesn't simultaneously contain pair of people (x, y) and pair of people (y, x).

Output

Print a single integer — the minimum sum of debts in the optimal rearrangement.

Examples

Input

5 3
1 2 10
2 3 1
2 4 1


Output

10


Input

3 0


Output

0


Input

4 3
1 2 1
2 3 1
3 1 1


Output

0

Note

In the first sample, you can assume that person number 1 owes 8 rubles to person number 2, 1 ruble to person number 3 and 1 ruble to person number 4. He doesn't owe anybody else anything. In the end, the total debt equals 10.

In the second sample, there are no debts.

In the third sample, you can annul all the debts.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskB solver = new TaskB();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskB {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        int v[] = new int[n];
        for(int i = 0;i < m; i++) {
           int a[] = in.parseInt1D(3);
           v[a[0]-1] -= a[2];
           v[a[1]-1] += a[2];
        }
        int total = 0;
        for(int i=0;i<n;i++) {
            if(v[i]>0){
                total += v[i];
            }
        }
        out.println(total) ;
    }
}

class InputReader {
    private BufferedReader br;
    private StringTokenizer st;
    public InputReader(InputStream in) {
        br=new BufferedReader(new InputStreamReader(in));
        try {
            st=new StringTokenizer(br.readLine());
        } catch (IOException ignored) {

        }
    }

    public void readLine() {
        try {
            st=new StringTokenizer(br.readLine());
        } catch (IOException e) {
            return;
        }
    }
    public int nextInt(){
        return Integer.parseInt(st.nextToken());
    }

    /**
     * Parse 1D array from current StringTokenizer
     */
    public int[] parseInt1D(int n){
        readLine();
        int r[]=new int[n];
        for(int i=0;i<n;i++){
            r[i]=nextInt();
        }
        return r;
    }



}