Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

The problem statement has recently been changed. View the changes.

×
E. A Museum Robbery

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputThere's a famous museum in the city where Kleofáš lives. In the museum, *n* exhibits (numbered 1 through *n*) had been displayed for a long time; the *i*-th of those exhibits has value *v*_{i} and mass *w*_{i}.

Then, the museum was bought by a large financial group and started to vary the exhibits. At about the same time, Kleofáš... gained interest in the museum, so to say.

You should process *q* events of three types:

- type 1 — the museum displays an exhibit with value
*v*and mass*w*; the exhibit displayed in the*i*-th event of this type is numbered*n*+*i*(see sample explanation for more details) - type 2 — the museum removes the exhibit with number
*x*and stores it safely in its vault - type 3 — Kleofáš visits the museum and wonders (for no important reason at all, of course): if there was a robbery and exhibits with total mass at most
*m*were stolen, what would their maximum possible total value be?

For each event of type 3, let *s*(*m*) be the maximum possible total value of stolen exhibits with total mass ≤ *m*.

Formally, let *D* be the set of numbers of all exhibits that are currently displayed (so initially *D* = {1, ..., n}). Let *P*(*D*) be the set of all subsets of *D* and let

Then, *s*(*m*) is defined as

Compute *s*(*m*) for each . Note that the output follows a special format.

Input

The first line of the input contains two space-separated integers *n* and *k* (1 ≤ *n* ≤ 5000, 1 ≤ *k* ≤ 1000) — the initial number of exhibits in the museum and the maximum interesting mass of stolen exhibits.

Then, *n* lines follow. The *i*-th of them contains two space-separated positive integers *v*_{i} and *w*_{i} (1 ≤ *v*_{i} ≤ 1 000 000, 1 ≤ *w*_{i} ≤ 1000) — the value and mass of the *i*-th exhibit.

The next line contains a single integer *q* (1 ≤ *q* ≤ 30 000) — the number of events.

Each of the next *q* lines contains the description of one event in the following format:

- 1
*v**w*— an event of type 1, a new exhibit with value*v*and mass*w*has been added (1 ≤*v*≤ 1 000 000, 1 ≤*w*≤ 1000) - 2
*x*— an event of type 2, the exhibit with number*x*has been removed; it's guaranteed that the removed exhibit had been displayed at that time - 3 — an event of type 3, Kleofáš visits the museum and asks his question

There will be at most 10 000 events of type 1 and at least one event of type 3.

Output

As the number of values *s*(*m*) can get large, output the answers to events of type 3 in a special format.

For each event of type 3, consider the values *s*(*m*) computed for the question that Kleofáš asked in this event; print one line containing a single number

where *p* = 10^{7} + 19 and *q* = 10^{9} + 7.

Print the answers to events of type 3 in the order in which they appear in the input.

Examples

Input

3 10

30 4

60 6

5 1

9

3

1 42 5

1 20 3

3

2 2

2 4

3

1 40 6

3

Output

556674384

168191145

947033915

181541912

Input

3 1000

100 42

100 47

400 15

4

2 2

2 1

2 3

3

Output

0

Note

In the first sample, the numbers of displayed exhibits and values *s*(1), ..., *s*(10) for individual events of type 3 are, in order:

The values of individual exhibits are *v*_{1} = 30, *v*_{2} = 60, *v*_{3} = 5, *v*_{4} = 42, *v*_{5} = 20, *v*_{6} = 40 and their masses are *w*_{1} = 4, *w*_{2} = 6, *w*_{3} = 1, *w*_{4} = 5, *w*_{5} = 3, *w*_{6} = 6.

In the second sample, the only question is asked after removing all exhibits, so *s*(*m*) = 0 for any *m*.

Codeforces (c) Copyright 2010-2020 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Oct/29/2020 08:29:24 (f2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|