100 1
1000000 100 1000
 5 3
1 4 10
2 2 5
3 1 6
 10 4
1 5 20
2 5 200
100 6 20
101 1 100
 5 3
1 3 4
4 3 4
6 4 1
 4 1
6 1 1
 8 6
1 3 20
4 2 1
6 6 5
9 1 1
15 3 6
21 8 8
 4 5
1 2 3
2 1 3
3 1 2
4 3 3
5 4 1
 8 4
1 3 2
2 3 100
10 6 20
11 5 20
 4 10
1 1 1
3 1 2
4 1 2
6 1 2
8 1 2
13 1 2
16 1 1
17 1 2
19 3 1
20 1 1
 5 3
1 4 4
4 2 2
5 5 2
 4 3
1 3 10
2 2 15
12 4 1
 5 4
1 1 4
2 4 4
5 2 20
11 5 2
 1 10
4 1 1
9 1 1
10 1 1
12 1 1
13 1 1
15 1 1
16 1 1
18 1 1
19 1 1
20 1 1

5050
 10
-1
5
 15
40
-1
1
 6
-1
10
 1
 6
9
-1
4
15
36
 3
3
4
-1
10
 6
15
-1
21
 1
1
2
1
1
1
1
1
6
1
 10
-1
15
 6
-1
10
 1
14
-1
15
 1
1
1
1
1
1
1
1
1
1

#include <bits/stdc++.h>
using namespace std;
long long n, q, curr, tme, m, tmp;
struct my {
  long long t, k, d, y;
} a[100100];
set<pair<long long, long long> > Q;
long long b[110];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 1; i <= q; i++) cin >> a[i].t >> a[i].k >> a[i].d;
  curr = n;
  int j = 1;
  for (int i = 1; i <= 1010000 && j <= n; i++) {
    while (i == Q.begin()->first && Q.size()) {
      for (int z = 1; z <= 100; z++)
        if (b[z] == Q.begin()->second) b[z] = 0;
      curr += a[Q.begin()->second].k,
          Q.erase({a[Q.begin()->second].t + a[Q.begin()->second].d,
                   Q.begin()->second});
    }
    if (a[j].t == i) {
      long long tmp = 0;
      if (curr >= a[j].k) {
        curr -= a[j].k, Q.insert({a[j].d + a[j].t, j});
        long long u = a[j].k;
        for (int z = 1; z <= 100 && u; z++)
          if (!b[z]) b[z] = j, tmp += z, u--;
        cout << tmp << "\n";
      } else
        cout << "-1\n";
      j++;
    }
  }
  return 0;
}
