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>
int main() {
  int n, m, count = 0, k, q, T[100000], K[100000], D[100000], O[100][1000], i,
            j, res[100000], count2 = 0, l;
  scanf("%d", &n);
  scanf("%d", &q);
  for (i = 1; i <= n; i++) {
    res[i] = 0;
    for (j = 1; j <= 1000; j++) {
      O[i][j] = 0;
    }
  }
  for (i = 1; i <= q; i++) {
    count = 0;
    count2 = 0;
    scanf("%d%d%d", &T[i], &K[i], &D[i]);
    for (k = 1; k <= n && count <= K[i]; k++) {
      if (O[k][T[i]] == 0) {
        count++;
      }
    }
    if (count >= K[i]) {
      for (j = 1; j <= n && count2 < K[i]; j++) {
        if (O[j][T[i]] == 0) {
          count2++;
          res[i] += l;
          for (m = T[i]; m < (D[i] + T[i]); m++) {
            O[l][m] = 1;
          }
        }
      }
    } else {
      res[i] = -1;
    }
  }
  for (i = 1; i <= q; i++) {
    printf("%d\n", res[i]);
  }
  return 0;
}
