# [Solution] Bananas in a Microwave Codeforces Round #711

Bananas in a Microwave Solution Codeforces

You have a malfunctioning microwave in which you want to put some bananas. You have nn time-steps before the microwave stops working completely. At each time-step, it displays a new operation.

Let kk be the number of bananas in the microwave currently. Initially, k=0k=0. In the ii-th operation, you are given three parameters titi, xixi, yiyi in the input. Based on the value of titi, you must do one of the following:

Type 1: (ti=1ti=1, xixi, yiyi) — pick an aiai, such that 0≤ai≤yi0≤ai≤yi, and perform the following update aiai times: k:=⌈(k+xi)⌉k:=⌈(k+xi)⌉.

Type 2: (ti=2ti=2, xixi, yiyi) — pick an aiai, such that 0≤ai≤yi0≤ai≤yi, and perform the following update aiai times: k:=⌈(k⋅xi)⌉k:=⌈(k⋅xi)⌉.

Note that xixi can be a fractional value. See input format for more details. Also, ⌈x⌉⌈x⌉ is the smallest integer ≥x≥x.

At the ii-th time-step, you must apply the ii-th operation exactly once.

For each jj such that 1≤j≤m1≤j≤m, output the earliest time-step at which you can create exactly jj bananas. If you cannot create exactly jj bananas, output −1−1.Input

The first line contains two space-separated integers nn (1≤n≤200)(1≤n≤200) and mm (2≤m≤105)(2≤m≤105).

Then, nn lines follow, where the ii-th line denotes the operation for the ii-th timestep. Each such line contains three space-separated integers titi, x′ixi′ and yiyi (1≤ti≤21≤ti≤2, 1≤yi≤m1≤yi≤m).

Note that you are given x′ixi′, which is 105⋅xi105⋅xi. Thus, to obtain xixi, use the formula xi=x′i105xi=xi′105.

For type 1 operations, 1≤x′i≤105⋅m1≤xi′≤105⋅m, and for type 2 operations, 105<x′i≤105⋅m105<xi′≤105⋅m.Output

Print mm integers, where the ii-th integer is the earliest time-step when you can obtain exactly ii bananas (or −1−1 if it is impossible).

Examples: Bananas in a Microwave Solution Codeforces

input

```3 20
1 300000 2
2 400000 2
1 1000000 3
```

output

```-1 -1 1 -1 -1 1 -1 -1 -1 3 -1 2 3 -1 -1 3 -1 -1 -1 3
```

input

```3 20
1 399999 2
2 412345 2
1 1000001 3
```

outputCopy

```-1 -1 -1 1 -1 -1 -1 1 -1 -1 3 -1 -1 -1 3 -1 2 -1 3 -1
```

Note

In the first sample input, let us see how to create 1616 number of bananas in three timesteps. Initially, k=0k=0.

• In timestep 1, we choose a1=2a1=2, so we apply the type 1 update — k:=⌈(k+3)⌉k:=⌈(k+3)⌉ — two times. Hence, kk is now 6.
• In timestep 2, we choose a2=0a2=0, hence value of kk remains unchanged.
• In timestep 3, we choose a3=1a3=1, so we are applying the type 1 update k:=⌈(k+10)⌉k:=⌈(k+10)⌉ once. Hence, kk is now 16.

It can be shown that k=16k=16 cannot be reached in fewer than three timesteps with the given operations.

In the second sample input, let us see how to create 1717 number of bananas in two timesteps. Initially, k=0k=0.

• In timestep 1, we choose a1=1a1=1, so we apply the type 1 update — k:=⌈(k+3.99999)⌉k:=⌈(k+3.99999)⌉ — once. Hence, kk is now 4.
• In timestep 2, we choose a2=1a2=1, so we apply the type 2 update — k:=⌈(k⋅4.12345)⌉k:=⌈(k⋅4.12345)⌉ — once. Hence, kk is now 17.

It can be shown that k=17k=17 cannot be reached in fewer than two timesteps with the given operations.

Solution: Coming Soon