# Root of the Problem

### Problem Statement :

```Chef has a binary tree. The binary tree consists of 1 or more nodes. Each node has a unique integer id. Each node has up to 2 children, which are identified by their ids, and each node is the child of at most 1 other node. A node X is considered to be an ancestor of node Y if node Y is a child of node X or if there is some node Z for which X is an ancestor of Z and Y is a child of Z. No node is an ancestor of itself. A special node called the root node is an ancestor of all other nodes.

Chef has forgotten which node of his tree is the root, and wants you to help him to figure it out. Unfortunately, Chef's knowledge of the tree is incomplete. He does not remember the ids of the children of each node, but only remembers the sum of the ids of the children of each node.

Input

Input begins with an integer T, the number of test cases. Each test case begins with an integer N, the number of nodes in the tree. N lines follow with 2 integers each: the id of a node, and the sum of the ids of its children. The second number will be 0 if the node has no children.

Output

For each test case, output on a line a space separated list of all possible values for the id of the root node in increasing order. It is guaranteed that at least one such id exists for each test case.

Constraints

1 ≤ T ≤ 50
1 ≤ N ≤ 30
All node ids are between 1 and 1000, inclusive

Sample Input

2
1
4 0
6
1 5
2 0
3 0
4 0
5 5
6 5

Sample Output

4
6```

### Solution :

```                            ```Solution in C :

#include <stdio.h>

int main(void) {
int t, n, id_sum, sid_sum, x, y;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
id_sum = 0, sid_sum = 0;
while(n--) {
scanf("%d %d", &x, &y);
id_sum += x;
sid_sum += y;
}
printf("%d\n", id_sum - sid_sum);
}
return 0;
}```
```

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

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

#define ll long long int

int main() {
ll t;
cin>>t;
for(ll z=1;z<=t;z++){
ll n;
cin>>n;
int ans=0;
for(int j=1;j<=n;j++){

int x,y;
cin>>x>>y;
ans+=(x-y);
}
cout<<ans<<'\n';
}
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 void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n = sc.nextInt();
int node = 0;
int idsum= 0;
for(int i=0;i<n;i++)
{
node = node + sc.nextInt();
idsum = idsum + sc.nextInt();
}
int r = node - idsum;
System.out.println(r);
}
}
}```
```

```                        ```Solution in Python :

# cook your dish here
t=int(input())
for _ in range(t):
n = int(input())
root = 0
for j in range(n):
ide,c = map(int,input().split(' '))
root +=ide - c

print(root)```
```

## 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.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F