#include <bits/stdc++.h>
using namespace std;
#define int long long
#define fi first
#define se second
#define all(a) a.begin(), a.end()
#define el '\n'
const int maxN = 505;
const int base = 311;
const int MOD = 1e9+7;
const int INF = 1e15;
const int NEG_INF = -1e15;
const int MAX_DAYS = 1000;
int n, m, k, q;
string S, T, SS[maxN];
// int dp[maxN][maxN];
vector<int> g[maxN];
void run_test() {
int limit;
int type_num;
std::cin >> limit >> type_num;
std::map<int, vector<pair<int, int>>> by_weight;
for (int t = 0; t < type_num; t++) {
int value;
int weight;
int amt;
std::cin >> value >> weight >> amt;
if (weight <= limit && amt > 0) { by_weight[weight].push_back({value, amt}); }
}
/*
* best[i][j] contains the most value we can
* get using j weight and the first i weight types
*/
vector<vector<long long>> best(by_weight.size() + 1,
vector<long long>(limit + 1, NEG_INF));
best[0][0] = 0;
int at = 1;
for (auto &[w, items] : by_weight) {
// sort items in reverse order by value
std::sort(items.begin(), items.end(), std::greater<pair<int, int>>());
for (int i = 0; i <= limit; i++) {
best[at][i] = best[at - 1][i];
int copies = 0;
int type_at = 0;
int curr_used = 0;
long long profit = 0;
// go through as many items until we run out of items or usable
// weight
while ((copies + 1) * w <= i && type_at < items.size()) {
copies++;
profit += items[type_at].first;
if (best[at - 1][i - copies * w] != NEG_INF) {
best[at][i] =
std::max(best[at][i], best[at - 1][i - copies * w] + profit);
}
curr_used++;
if (curr_used == items[type_at].second) {
curr_used = 0;
type_at++;
}
}
}
at++;
}
cout << *std::max_element(best.back().begin(), best.back().end()) << endl;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
// freopen("snakes.in", "r", stdin);
// freopen("snakes.out", "w", stdout);
int test = 1;
// cin >> test;
while (test-- > 0)
{
run_test();
}
}
