description: Tanya is learning how to add numbers, but so far she is not doing it correctly. She is adding two numbers a and b using the following algorithm:
  
    1. If one of the numbers is shorter than the other, Tanya adds leading zeros so that the numbers are the same length. 
    2. The numbers are processed from right to left (that is, from the least significant digits to the most significant). 
    3. In the first step, she adds the last digit of a to the last digit of b and writes their sum in the answer. 
    4. At each next step, she performs the same operation on each pair of digits in the same place and writes the result to the left side of the answer. 
  
  
  
  For example, the numbers a = 17236 and b = 3465 Tanya adds up as follows:
  
  $$$ \large{ \begin{array}{r} + \begin{array}{r} 17236\\\ 03465\\\ \end{array} \\\ \hline \begin{array}{r} 1106911 \end{array} \end{array}} $$$
  
    * calculates the sum of 6 + 5 = 11 and writes 11 in the answer. 
    * calculates the sum of 3 + 6 = 9 and writes the result to the left side of the answer to get 911. 
    * calculates the sum of 2 + 4 = 6 and writes the result to the left side of the answer to get 6911. 
    * calculates the sum of 7 + 3 = 10, and writes the result to the left side of the answer to get 106911. 
    * calculates the sum of 1 + 0 = 1 and writes the result to the left side of the answer and get 1106911. 
  
  
  
  As a result, she gets 1106911.
  
  You are given two positive integers a and s. Find the number b such that by adding a and b as described above, Tanya will get s. Or determine that no suitable b exists.
  
  Input
  
  The first line of input data contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  Each test case consists of a single line containing two positive integers a and s (1 ≤ a < s ≤ 10^{18}) separated by a space.
  
  Output
  
  For each test case print the answer on a separate line.
  
  If the solution exists, print a single positive integer b. The answer must be written without leading zeros. If multiple answers exist, print any of them.
  
  If no suitable number b exists, output -1.
  
  Example
  
  Input
  
  
  6
  17236 1106911
  1 5
  108 112
  12345 1023412
  1 11
  1 20
  
  
  Output
  
  
  3465
  4
  -1
  90007
  10
  -1
  
  Note
  
  The first test case is explained in the main part of the statement.
  
  In the third test case, we cannot choose b that satisfies the problem statement.
solution: //package currentContest;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  
  public class P1 {
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
  		StringBuilder st=new StringBuilder();
  		while(t--!=0){
  			long a=sc.nextLong();
  			long s=sc.nextLong();
  			int flag=0;
  			long b=0;
  			long mult=1;
  			while(s!=0) {
  				long last=s%10;
  				long toadd=a%10;
  				s=s/10;
  				a=a/10;
  				if(toadd==last) {
  					mult*=10;
  				}else if(toadd<last) {
  					b=b+(last-toadd)*mult;
  					mult*=10;
  				}else {
  					b=b+(10-toadd+last)*mult;
  					mult*=10;
  					if(s%10!=1) {
  						flag=-1;
  						break;
  					}
  					s=s/10;
  				}
  //				System.out.println(b);
  			}
  			if(a!=0||flag==-1) {
  				st.append(-1+"\n");
  			}else {
  				st.append(b+"\n");
  			}
  		}
  		System.out.println(st);
  	}
  	
  	static class DisjointUnionSets {
  		int[] rank, parent;
  		int n;
  		
  		// Constructor
  		public DisjointUnionSets(int n)
  		{
  			rank = new int[n];
  			parent = new int[n];
  			this.n = n;
  			makeSet();
  		}
  		
  		// Creates n sets with single item in each
  		void makeSet()
  		{
  			for (int i = 0; i < n; i++) {
  				// Initially, all elements are in
  				// their own set.
  				parent[i] = i;
  			}
  		}
  		
  		// Returns representative of x's set
  		int find(int x)
  		{
  			// Finds the representative of the set
  			// that x is an element of
  			if (parent[x] != x) {
  				// if x is not the parent of itself
  				// Then x is not the representative of
  				// his set,
  				parent[x] = find(parent[x]);
  				
  				// so we recursively call Find on its parent
  				// and move i's node directly under the
  				// representative of this set
  			}
  			
  			return parent[x];
  		}
  		
  		// Unites the set that includes x and the set
  		// that includes x
  		void union(int x, int y)
  		{
  			// Find representatives of two sets
  			int xRoot = find(x), yRoot = find(y);
  			
  			// Elements are in the same set, no need
  			// to unite anything.
  			if (xRoot == yRoot)
  				return;
  			
  			// If x's rank is less than y's rank
  			if (rank[xRoot] < rank[yRoot])
  				
  				// Then move x under y  so that depth
  				// of tree remains less
  				parent[xRoot] = yRoot;
  			
  			// Else if y's rank is less than x's rank
  			else if (rank[yRoot] < rank[xRoot])
  				
  				// Then move y under x so that depth of
  				// tree remains less
  				parent[yRoot] = xRoot;
  			
  			else // if ranks are the same
  			{
  				// Then move y under x (doesn't matter
  				// which one goes where)
  				parent[yRoot] = xRoot;
  				
  				// And increment the result tree's
  				// rank by 1
  				rank[xRoot] = rank[xRoot] + 1;
  			}
  		}
  	}
  
  	static FastReader sc = new FastReader();
  
  	public static void solvegraph() {
  		int n = sc.nextInt();
  
  		int edge[][] = new int[n - 1][2];
  		for (int i = 0; i < n - 1; i++) {
  			edge[i][0] = sc.nextInt() - 1;
  			edge[i][1] = sc.nextInt() - 1;
  		}
  		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
  		for (int i = 0; i < n; i++) {
  			ad.add(new ArrayList<Integer>());
  		}
  		for (int i = 0; i < n - 1; i++) {
  			ad.get(edge[i][0]).add(edge[i][1]);
  			ad.get(edge[i][1]).add(edge[i][0]);
  		}
  		int parent[] = new int[n];
  		Arrays.fill(parent, -1);
  		parent[0] = n;
  		ArrayDeque<Integer> queue = new ArrayDeque<>();
  		queue.add(0);
  		int child[] = new int[n];
  		Arrays.fill(child, 0);
  		ArrayList<Integer> lv = new ArrayList<Integer>();
  		while (!queue.isEmpty()) {
  			int toget = queue.getFirst();
  			queue.removeFirst();
  			child[toget] = ad.get(toget).size() - 1;
  			for (int i = 0; i < ad.get(toget).size(); i++) {
  				if (parent[ad.get(toget).get(i)] == -1) {
  					parent[ad.get(toget).get(i)] = toget;
  					queue.addLast(ad.get(toget).get(i));
  				}
  			}
  			lv.add(toget);
  		}
  
  		child[0]++;
  	}
  
  	static void sort(int[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			int tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  	static void sort(char[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			char tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static void sort(long[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			long tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static String sort(String s) {
  		Character ch[] = new Character[s.length()];
  		for (int i = 0; i < s.length(); i++) {
  			ch[i] = s.charAt(i);
  		}
  		Arrays.sort(ch);
  		StringBuffer st = new StringBuffer("");
  		for (int i = 0; i < s.length(); i++) {
  			st.append(ch[i]);
  		}
  		return st.toString();
  	}
  
  	public static long gcd(long a, long b) {
  		if (a == 0) {
  			return b;
  		}
  		return gcd(b % a, a);
  	}
  
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  }
  