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.*;
import java.util.*;

public class Main {public static void main(String[] args) throws Exception {new Solve();}}

class Solve { public Solve() throws Exception {solve();}
    static BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st = new StringTokenizer("");

    void solve() throws Exception {
        int n, m;
        n = NI();
        m = NI();
        int t, q, w;
        Pair debt[] = new Pair[n];
        for (int i = 0; i < n; i++)
            debt[i] = new Pair();
        for (int i = 0; i < m; i++) {
            t = NI();
            q = NI();
            w = NI();
            debt[t - 1].a += w;
            debt[q - 1].b += w;
        }
        int res = 0;
        for (Pair p : debt) {
            t = p.b - p.a;
            res += t > 0 ? t : 0;
        }
        System.out.println(res);
    }


    class Pair {
        int a, b;
        public Pair() {
           // i = q;
            a = 0;
            b = 0;
        }
    }

    int min(int i1, int i2) {return i1 < i2 ? i1 : i2;}

    long min(long i1, long i2) {return i1 < i2 ? i1 : i2;}

    int max(int i1, int i2) {return i1 > i2 ? i1 : i2;}

    long max(long i1, long i2) {return i1 > i2 ? i1 : i2;}

    public String NS() throws Exception {
        while (!st.hasMoreTokens()) st = new StringTokenizer(stdin.readLine());
        return st.nextToken();}

    public int NI() throws Exception {return Integer.parseInt(NS());}

    public long NL() throws Exception {return Long.parseLong(NS());}

    int abs(int x) {return x < 0 ? -x : x;}

}