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() {
  int n, q, p[1000009], oc[101] = {0};
  cin >> n >> q;
  int fre = n, x;
  pair<int, int> que[1000009];
  for (int i = (0); i < (1000009); ++i) {
    que[i].first = 0, que[i].second = 0;
  }
  for (int i = (0); i < (q); ++i) {
    scanf("%d", &x);
    scanf("%d %d", &que[x].first, &que[x].second);
  }
  for (int i = (1); i < (x + 5); ++i) {
    if (p[i] != 0) {
      fre += p[i];
      int cnt = 0;
      for (int q = (1); q < (n + 1); ++q)
        if (oc[q] == i) {
          oc[q] = 0;
        }
      p[i] = 0;
    }
    if (que[i].first != 0) {
      if (fre >= que[i].first) {
        fre -= que[i].first;
        int sum = 0;
        int temp = que[i].first;
        for (int j = (1); j < (n + 1); ++j) {
          if (!temp) break;
          if (!oc[j]) {
            sum += j;
            oc[j] = i + que[i].second;
            temp--;
          }
        }
        p[i + que[i].second] = que[i].first;
        cout << sum << endl;
      } else {
        cout << -1 << endl;
      }
    }
  }
}
