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;
template <typename T, typename U>
inline void amin(T &x, U &y) {
  if (y < x) swap(x, y);
}
template <typename T, typename U>
inline void amax(T &x, U &y) {
  if (x < y) swap(x, y);
}
long long int t[100005], k[100005], d[100005], tu[1010000] = {0}, ans[100005];
int main() {
  ios_base::sync_with_stdio(false);
  long long int n, q, s[101] = {0};
  cin >> n >> q;
  for (int i = 0; i < q; i++) cin >> t[i] >> k[i] >> d[i];
  for (int i = 0; i < q; i++) {
    if (tu[t[i]] + k[i] > n)
      ans[i] = -1;
    else {
      for (int j = 1; j <= n; j++)
        if (s[j] < t[i]) s[j] = 0;
      long long int l = 1;
      for (int j = 1; j <= k[i];) {
        if (!s[l]) {
          ans[i] += l;
          s[l] = t[i] + d[i] - 1;
          j++;
        } else {
          l++;
        }
      }
      for (long long int j = t[i]; j < t[i] + d[i]; j++) tu[j] += k[i];
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
  return 0;
}
