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

B. DZY Loves FFT

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputDZY loves Fast Fourier Transformation, and he enjoys using it.

Fast Fourier Transformation is an algorithm used to calculate convolution. Specifically, if *a*, *b* and *c* are sequences with length *n*, which are indexed from 0 to *n* - 1, and

We can calculate *c* fast using Fast Fourier Transformation.

DZY made a little change on this formula. Now

To make things easier, *a* is a permutation of integers from 1 to *n*, and *b* is a sequence only containing 0 and 1. Given *a* and *b*, DZY needs your help to calculate *c*.

Because he is naughty, DZY provides a special way to get *a* and *b*. What you need is only three integers *n*, *d*, *x*. After getting them, use the code below to generate *a* and *b*.

//x is 64-bit variable;

function getNextX() {

x = (x * 37 + 10007) % 1000000007;

return x;

}

function initAB() {

for(i = 0; i < n; i = i + 1){

a[i] = i + 1;

}

for(i = 0; i < n; i = i + 1){

swap(a[i], a[getNextX() % (i + 1)]);

}

for(i = 0; i < n; i = i + 1){

if (i < d)

b[i] = 1;

else

b[i] = 0;

}

for(i = 0; i < n; i = i + 1){

swap(b[i], b[getNextX() % (i + 1)]);

}

}

Operation x % y denotes remainder after division *x* by *y*. Function swap(x, y) swaps two values *x* and *y*.

Input

The only line of input contains three space-separated integers *n*, *d*, *x* (1 ≤ *d* ≤ *n* ≤ 100000; 0 ≤ *x* ≤ 1000000006). Because DZY is naughty, *x* can't be equal to 27777500.

Output

Output *n* lines, the *i*-th line should contain an integer *c*_{i - 1}.

Examples

Input

3 1 1

Output

1

3

2

Input

5 4 2

Output

2

2

4

5

5

Input

5 4 3

Output

5

5

5

5

4

Note

In the first sample, *a* is [1 3 2], *b* is [1 0 0], so *c*_{0} = *max*(1·1) = 1, *c*_{1} = *max*(1·0, 3·1) = 3, *c*_{2} = *max*(1·0, 3·0, 2·1) = 2.

In the second sample, *a* is [2 1 4 5 3], *b* is [1 1 1 0 1].

In the third sample, *a* is [5 2 1 4 3], *b* is [1 1 1 1 0].

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Oct/23/2019 22:31:54 (f3).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|