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;
int main() {
  ios::sync_with_stdio(0);
  long long n, t, k;
  cin >> n >> t;
  vector<long long> b(n), ns(t), be(t), d(t), ku(t);
  set<long long> gq;
  for (long long i = 0; i < (long long)t; i++) {
    cin >> be[i] >> ku[i] >> d[i];
    while (!gq.empty() && *gq.begin() <= be[i]) {
      k = *gq.begin();
      for (long long j = 0; j < (long long)n; j++)
        if (b[j] == k) b[j] = 0;
      gq.erase(gq.begin());
    }
    gq.insert(be[i] + d[i]);
    long long c = 0;
    for (long long j = 0; j < (long long)n; j++)
      if (!b[j]) {
        if (ku[i] == 0) break;
        b[j] = be[i] + d[i];
        c += j + 1;
        ku[i]--;
      }
    if (ku[i])
      ns[i] = -1;
    else
      ns[i] = c;
  }
  for (long long qz = 0; qz < (long long)ns.size(); qz++)
    cout << (ns)[qz] << ' ';
  cout << "\n";
  return 0;
}
