# 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
{
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 :

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)```
```

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio