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 N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
int t[N], k[N], d[N];
int vis[105];
int n, q, i, j;
int main() {
  scanf("%d%d", &n, &q);
  for (i = 1; i <= q; ++i) {
    int cnt = 0;
    scanf("%d%d%d", &t[i], &k[i], &d[i]);
    for (j = 1; j <= n; ++j) {
      if (!vis[j] || d[vis[j]] + t[vis[j]] <= t[i]) cnt++;
    }
    if (cnt < k[i]) {
      printf("-1\n");
      continue;
    }
    cnt = 0;
    int s = 0;
    for (j = 1; j <= n && cnt < k[i]; ++j) {
      if (!vis[j] || d[vis[j]] + t[vis[j]] <= t[i]) {
        vis[j] = i;
        cnt++;
        s += j;
      }
    }
    printf("%d\n", s);
  }
  return 0;
}
