# Marbles

### Problem Statement :

```Rohit dreams he is in a shop with an infinite amount of marbles. He is allowed to select n marbles. There are marbles of k different colors. From each color there are also infinitely many marbles. Rohit wants to have at least one marble of each color, but still there are a lot of possibilities for his selection. In his effort to make a decision he wakes up. Now he asks you how many possibilities for his selection he would have had. Assume that marbles of equal color can't be distinguished, and the order of the marbles is irrelevant.

Input

The first line of input contains a number T <= 100 that indicates the number of test cases to follow. Each test case consists of one line containing n and k, where n is the number of marbles Rohit selects and k is the number of different colors of the marbles. You can assume that 1<=k<=n<=1000000.

Output

For each test case print the number of possibilities that Rohit would have had. You can assume that this number fits into a signed 64 bit integer.

Example

Input:

2
10 10
30 7

Output:

1
475020```

### Solution :

```                            ```Solution in C :

#include <stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
long long int n,k;scanf("%lld %lld",&n,&k);
long long int d=n-k,sum=1;
if(n==k) {
sum=1;
}
else
{
for(long long int i=1;i<k;i++)
{
sum=sum*(d+i);
sum=sum/i;
}
}
printf("%lld\n",sum);
}
return 0;
}```
```

```                        ```Solution in C++ :

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define  ld long double
#define vl vector<ll>
#define vi vector<int>
#define vs vector<string>
#define vvl vector<vl>
#define vvvl vector<vvl>
#define  vvs vector<vs>
#define   vvi vector<vi>
#define  vs vector<string>
#define  vvs vector<vs>
using namespace std;

ll comb(ll n, ll k){
ll ans = 1;
if (2*k > n) k = n-k;

for(ll i=1; i<=k; i++){
ans *= n;
ans /= i;
n--;
}

return ans;
}

void solve(){
ll n,k; cin>>n>>k;

cout << comb(n-1, k-1) << endl;
}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll t;
cin>>t;
while(t--)
solve();

return 0;
}```
```

```                        ```Solution in Java :

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static long fact(int n)
{
long res=0;
if(n==1||n==0)
{
res= 1;
}
else{
res =  n*fact(n-1);
}
return res;
}
public static void main (String[] args) throws java.lang.Exception
{
int n,k,t;
long poss;
Scanner sc = new Scanner(System.in);
t = sc.nextInt();
while(t>0)
{
t--;
n = sc.nextInt();
k = sc.nextInt();
if(n==k)
{
poss =1;
}
else
{
int m = n-k; poss = 1;
for(int i=1;i<k;i++)
{
poss = poss*(m+i)/i;
}
}
System.out.println(poss);
}
}
}```
```

```                        ```Solution in Python :

n=int(input())
for i in range(n):
n,b=map(int,input().split())
r1=1
for i in range(1,b):
r1=r1*(n-b+i)//(i)
print(int(r1))```
```

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.