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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

typedef long long ll;
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define all(x) x.begin(), x.end()
#define f(i,a,b) for(int i = (a); i <= (b); i++)
#define fd(i,a,b) for(int i = (a); i >= (b); i--)
#define mp make_pair
#define faster_io() ios_base::sync_with_stdio(false)
#define pb push_back
#define pii pair<int,int>
#define SZ(x) ((int)x.size())
#define vii vector<pair<int,int>>

const int INF = 2000000005;
const ll INFLL = 1000000000000000002ll;
const ll MOD = 1000000007;

inline ll min(ll a, ll b, ll c){return min(min(a,b),c);}
inline ll min(ll a, ll b, ll c, ll d){return min(min(min(a,b),c),d);}
inline ll max(ll a, ll b, ll c){return max(max(a,b),c);}
inline ll max(ll a, ll b, ll c, ll d){return max(max(max(a,b),c),d);}

// -------------------------------------------------------------------------------------------------------------

int Free[1002005], Mark[1002005], S[105], Task[1002005], N, Q, Ans[100005];
pii Ev[1002005];

int main()
{
   cin >> N >> Q;
   f(i,1,100) S[i] = S[i-1] + i;
   int left = 1;
   f(q,1,Q)
   {
       int t,k,d;
       scanf("%d %d %d", &t, &k, &d);
       Ev[t].first = k;
       Ev[t].second = d;
       Task[t] = q;
   }
   f(i,1,1000000)
   {
       left -= Free[i];
       if(N-left+1 < Ev[i].first)
       {
           Ans[Task[i]] = -1;
           continue;
       }
       Ans[Task[i]] = S[left + Ev[i].first - 1] - S[left-1];
       Free[i + Ev[i].second] += Ev[i].first;
       left += Ev[i].first;
   }
   f(i,1,Q) cout << Ans[i] << "\n";
}
