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;
using ll = long long;
using pll = pair<long long, long long>;
using pii = pair<int, int>;
using vi = vector<int>;
const int N = 100005;
int t[N], k[N], d[N];
int tim[N];
set<int> fr;
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d%d", &t[i], &k[i], &d[i]);
    int inter = t[i] - t[i - 1];
    int ans = 0;
    for (int j = 1; j <= n; ++j) {
      tim[j] = max(0, tim[j] - inter);
      if (tim[j] == 0) fr.insert(j);
    }
    if (fr.size() >= k[i]) {
      ans = 0;
      int need = k[i];
      for (set<int>::iterator it = fr.begin(); it != fr.end();) {
        if (need == 0) break;
        tim[*it] = d[i];
        ans += (*it);
        need--;
        fr.erase(it++);
      }
    } else {
      ans = -1;
    }
    printf("%d\n", ans);
  }
  return 0;
}
