Deep in the countryside, Farmer John’s cows aren’t just ordinary farm animals—they are part of a clandestine bovine intelligence network. Each cow carries an ID number, carefully assigned by the elite cow cryptographers. However, due to Farmer John's rather haphazard tagging system, some cows ended up with the same ID.

Farmer John noted that there are $N$ ($1\le N\le 2\cdot 10^5$) unique ID numbers, and for each unique ID $d_i$ ($0\le d_i\le 10^9$), there are $n_i$ ($1\le n_i\le 10^9$) cows who shared it.

The cows can only communicate in pairs, and their secret encryption method has one strict rule: two cows can only exchange information if they are not the same cow and the sum of their ID numbers equals either $A$ or $B$ ($0\le A\le B\le 2\cdot 10^9$). A cow can only engage in one conversation at a time (i.e., no cow can be part of more than one pair).

Farmer John would like to maximize the number of disjoint communication pairs to ensure the best information flow. Can you determine the largest number of conversations that can happen at once?

SAMPLE INPUT:

4 4 5
17 2
100 0
10 1
200 4

SAMPLE OUTPUT:

118

A cow with an ID of $4$ can also communicate with another cow with an ID of $1$ (sum to $5$). There are $10$ cows of ID $1$ and $100$ remaining unpaired cows of ID $4$, allowing for another $10$ pairs.

Finally, a cow with an ID of $2$ can communicate with another cow of the same ID. Since there are a total of $17$ cows of ID $2$, there can be up to $8$ more pairs.

In total, there are $100+10+8=118$ communicating pairs. It can be shown that this is the maximum possible number of pairs.

SAMPLE INPUT:

4 4 5
100 0
10 1
100 3
20 4

SAMPLE OUTPUT:

30

Pairing IDs $0$ and $4$ makes $20$ pairs, while pairing IDs $1$ and $3$ makes $10$ pairs. It can be shown that this is the optimal pairing, resulting in a total of $30$ pairs.

Problem credits: Benjamin Qi