# The Legend Was Invented, but Not the Name

You are given two numbers $x$ and $y$. You need to find an array of numbers such that the sum of the squares of these numbers is equal to $10_{15}+x$, and the sum of the cubes is equal to $10_{15}+y$. Formally, you need to find any array of numbers such that:

where $m$ is the number of elements in this array.

## Input

The first and only line contains two integers $x$ and $y$ ($0≤x,y≤10_{9}$).

## Output

In the first line, you need to output only one integer $n$ $(1≤n≤2⋅10_{5})$ — the number of distinct elements in the array $a$.

In the next $n$ lines you need to output the information about the array in the following form:

$c_{i}k_{i}$

That means that there are $k_{i}$ elements in the array that are equal to $c_{i}$ $(1≤i≤n,1≤c_{i}≤10_{9},1≤k_{i}≤10_{16})$.

All $c_{i}$'s need to be pairwise distinct.

If there are multiple answers, you can output any of them.

If there are no needed arrays, you need to output $−1$.

## Examples

## Note

In the first example, there are $10_{15}−3$ elements equal to $1$ and $1$ element equal to $2$. This way, the sums are following:

$(10_{15}−3)⋅1_{2}+1⋅2_{2}=10_{15}+1$

$(10_{15}−3)⋅1_{3}+1⋅2_{3}=10_{15}+5$

In the second example, there are $10_{15}+1$ $1$'s and $1$ entry of each number from $2$ to $5$:

$(10_{15}+1)⋅1_{2}+1⋅2_{2}+1⋅3_{2}+1⋅4_{2}+1⋅5_{2}=(10_{15}+1)+4+9+16+25=10_{15}+55$

$(10_{15}+1)⋅1_{3}+1⋅2_{3}+1⋅3_{3}+1⋅4_{3}+1⋅5_{3}=(10_{15}+1)+8+27+64+125=10_{15}+225$

In the third sample, it can be shown that there are no arrays that would satisfy the conditions.

## Scoring

($12$ points): $x=y$;

($13$ points): $x>y$;

($76$ points): $x,y≤99999$;

($99$ points): no additional constraints.