#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <climits>

using namespace std;

int main() {
    int N, M; cin >> N >> M;
    // v contains the initial balances for each person.
    vector<int> v(N,0), w;
    for(int i = 0; i < M; i++) {
        int a, b, p;
        cin >> a >> b >> p;
        v[a] -= p;
        v[b] += p;
    }

    // The vector zero_sets contains all bitmasks for groups whose total balance is 0.
    vector<int> zero_sets;
    for(int i = 0; i < (1 << v.size()); i++) {
        int sum = 0;
        for(int j = 0; j < (int)v.size(); j++) {
            if(i & (1 << j)) {
                sum += v[j];
            }
        }
        if(sum == 0)
            zero_sets.push_back(i);
    }

    // For each element of zero_sets, find the largest number of groups with
    // zero balance it can be split into.
    int S = zero_sets.size();
    vector<int> groups(S,0);
    for(int i = 0; i < S; i++) {
        for(int j = 0; j < i; j++) {
            if(!(zero_sets[j] & ~zero_sets[i])) { // If j is a submask of i...
                groups[i] = max(groups[i], groups[j] + 1);
            }
        }
    }

    // Finally, correct for the pairs and singles we found at the beginning.
    cout << N - groups[S - 1] << endl;
    return 0;
}