Problem9735--ABC195 —— D - Shipping Center

9735: ABC195 —— D - Shipping Center

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Time Limit : 1.000 sec  Memory Limit : 512 MiB

Description

We have $N$ pieces of baggage called Baggage $1$ through $N$, and $M$ boxes called Box $1$ through $M$.

Baggage $i$ has a size of $W_i$ and a value of $V_i$.

Box $i$ can contain a piece of baggage whose size of at most $X_i$. It cannot contain two or more pieces of baggage.

You will be given $Q$ queries. In each query, given two integers $L$ and $R$, solve the following problem:

-   Problem: Out of the $M$ boxes, $R-L+1$ boxes, Box $L,L+1,\ldots,R$, have become unavailable. Find the maximum possible total value of a set of baggage that we can put into the remaining boxes simultaneously.

Input

Input is given from Standard Input in the following format:

```
$N$ $M$ $Q$
$W_1$ $V_1$
$\vdots$
$W_N$ $V_N$
$X_1$ $\ldots$ $X_M$
$\mathrm{Query}_1$
$\vdots$
$\mathrm{Query}_Q$
```

Each Query is in the following format:

```
$L$ $R$
```

Output

Print $Q$ lines.

The $i$-th line should contain the answer to the problem described by $\mathrm{Query}_i$.

Constraints

-   $1 \leq N \leq 50$
-   $1 \leq M \leq 50$
-   $1 \leq Q \leq 50$
-   $1 \leq W_i \leq 10^6$
-   $1 \leq V_i \leq 10^6$
-   $1 \leq X_i \leq 10^6$
-   $1 \leq L \leq R \leq M$
-   All values in input are integers.

Sample 1 Input

3 4 3
1 9
5 3
7 8
1 8 6 9
4 4
1 4
1 3

Sample 1 Output

20
0
9

In the 1-st query, only Box 4 is unavailable. By putting Baggage 1 into Box 1, Baggage 3 into Box 2, and Baggage 2 into Box 3, we can put all baggage into boxes, making the total value of baggage in boxes 20.

In the 2-nd query, all boxes are unavailable; the answer is 0.

In the 3-rd query, only Box 4 is available. By putting Baggage 1 into Box 4, we can make the total value of baggage in boxes 9, which is the maximum possible result.

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