## Fetch the Treasure Codeforces Round #185 Div．2 (CF赛制功能测试)

Rainbow built h cells in a row that are numbered from 1 to h from left to right. There are n cells with treasure. We call each of these n cells "Treasure Cell". The i-th "Treasure Cell" is the ai-th cell and the value of treasure in it is ci dollars.

**LYD画了一行h个格子，编号为1到h。其中n个格子里有宝箱。第i个宝箱放在第ai个格子里，里面有ci大洋。**

Then, Freda went in the first cell. For now, she can go just k cells forward, or return to the first cell. That means Freda was able to reach the 1st, (k + 1)-th, (2•k + 1)-th, (3•k + 1)-th cells and so on.

**SHY来到了第一个格子。起初，她每次只能往前走k格，或者返回起点。这意味着SHY可以到达第1、k+1、2k+1、3k+1……等等个格子。**

Then Rainbow gave Freda m operations. Each operation is one of the following three types:

1. Add another method x: she can also go just x cells forward at any moment. For example, initially she has only one method k. If at some moment she has methods a1, a2, ..., ar then she can reach all the cells with number in form 1+∑(i=1~r) vi*ai , where vi — some non-negative integer.

**接下来有m个操作，操作有三种类型：
1. SHY 新技能get√ —— 她每次也可以往前走x个格子！**

2. Reduce the value of the treasure in the x-th "Treasure Cell" by y dollars. In other words, to apply assignment cx = cx - y.

**2. 第x个宝箱里少了y大洋 —— 可能被SJY吃♂了**

3. Ask the value of the most valuable treasure among the cells Freda can reach. If Freda cannot reach any cell with the treasure then consider the value of the most valuable treasure equal to 0, and do nothing. Otherwise take the most valuable treasure away. If several "Treasure Cells" have the most valuable treasure, take the "Treasure Cell" with the minimum number (not necessarily with the minimum number of cell). After that the total number of cells with a treasure is decreased by one.

As a programmer, you are asked by Freda to write a program to answer each query.

**3. 询问SHY能走到的哪个宝箱大洋最多，并把该宝箱里的大洋都拿走。不能到达任何宝箱输出0。多解选择能走到的编号最小的宝箱（请注意是宝箱编号最小，不一定是格子编号最小）。**

**作为一个码农，请你帮SHY写个程序回答每个询问。**

**Input**

The first line of the input contains four integers: h (1 ≤ h ≤ 10^18), n, m (1 ≤ n, m ≤ 10^5) and k (1 ≤ k ≤ 10^4).

Each of the next n lines contains two integers: ai (1 ≤ ai ≤ h), ci (1 ≤ ci ≤ 10^9). That means the i-th "Treasure Cell" is the ai-th cell and cost of the treasure in that cell is ci dollars. All the ai are distinct.

Each of the next m lines is in one of the three following formats:

• "1 x" — an operation of type 1, 1 ≤ x ≤ h;

• "2 x y" — an operation of type 2, 1 ≤ x ≤ n, 0 ≤ y < cx;

• "3" — an operation of type 3.

__ There are at most 20 operations of type 1__. It's guaranteed that at any moment treasure in each cell has positive value. It's guaranteed that all operations is correct (no operation can decrease the value of the taken treasure).

Please, do not use the %I64d specifier to read 64-bit integers in С++. It is preferred to use the

**cin, cout streams or the %lld specifier.**

**Output**

For each operation of type 3, output an integer indicates the value (in dollars) of the most valuable treasure among the "Treasure Cells" Freda can reach. If there is no such treasure, output 0.

**Sample test(s)
Input**

10 3 5 2

5 50

7 60

8 100

2 2 5

3

1 3

3

3

**Output**

55

100

50

**Note**

In the sample, there are 10 cells and 3 "Treasure Cells". The first "Treasure Cell" is cell 5, having 50 dollars tresure in it. The second "Treasure Cell" is cell 7, having 60 dollars tresure in it. The third "Treasure Cell" is cell 8, having 100 dollars tresure in it.

At first, Freda can only reach cell 1, 3, 5, 7 and 9. In the first operation, we reduce the value in the second "Treasure Cell" from 60 to 55. Then the most valuable treasure among the "Treasure Cells" she can reach is max(50, 55) = 55. After the third operation, she can also go 3 cells forward each step, being able to reach cell 1, 3, 4, 5, 6, 7, 8, 9, 10. So the most valuable tresure is 100.

Noticed that she took the 55 dollars and 100 dollars treasure away, so the last answer is 50.