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;
const int maxn = 200, maxq = 100100;
struct query {
  int t, k, d;
};
bool operator<(query a, query b) { return a.t < b.t; }
int n, q;
int free_time[maxn];
query queries[maxq];
int ok[maxn], ok_top;
int res[maxq];
int main(int argc, char** argv) {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) free_time[i] = 0;
  for (int i = 1; i <= q; i++)
    scanf("%d%d%d", &queries[i].t, &queries[i].k, &queries[i].d);
  sort(queries + 1, queries + 1 + q);
  for (int idx = 1; idx <= q; idx++) {
    int t = queries[idx].t, k = queries[idx].k, d = queries[idx].d;
    ok_top = 0;
    for (int i = 1; i <= n && ok_top < k; i++)
      if (free_time[i] < t) ok[++ok_top] = i;
    if (ok_top < k) {
      res[idx] = -1;
      continue;
    }
    int sum = 0;
    for (int i = 1; i <= ok_top; i++) {
      sum += ok[i];
      free_time[ok[i]] = t + d - 1;
    }
    res[idx] = sum;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", res[i]);
  return 0;
}
