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: import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.*;
  import java.io.*;
  public class Main {
      // Graph
      // prefix sums
      //inputs
      public static void main(String args[])throws Exception{
          Input sc=new Input();
          precalculates p=new precalculates();
          StringBuilder sb=new StringBuilder();
          int t=sc.readInt();
          for(int f=0;f<t;f++){
              long d[]=sc.readArrayLong();
              long a=d[0],s=d[1];
              long b=0;
              long temp=a;
              long sum=s;
              long k=1;
              while(a!=0){
                  //System.out.println(a+" "+s+" "+b);
                  if(a%10<=s%10){
                      b=b+((s%10-a%10))*k;
                      a/=10;
                      s/=10;
                  }else if(a%10>s%10){
                      b=b+(s%100-a%10)*k;
                      a/=10;
                      s/=100;
                  }
                  k*=10;
                  if(a==0 && s!=0){
                      b=b+(s)*k;
                  }
              }
              StringBuilder ss=new StringBuilder();
              long bb=b;
              boolean check=true;
              a=temp;
              b=bb;
              a=Math.min(b,a);
              b=Math.max(bb,temp);
              while(a!=0 && b!=0){
  
                      ss.insert(0,(a%10+b%10)+"");
                      a/=10;b/=10;
                 if(a==0 && b!=0)
                 {
                     ss.insert(0,b+"");
                 }
              }
              //System.out.println(temp+" "+bb+" "+ss+" "+sum);
              if(ss.toString().equals(""+sum)){
                  check=true;
              }else
                  check=false;
              if(check)
              sb.append(bb+"\n");
              else
                  sb.append("-1\n");
          }
          System.out.print(sb);
      }
  }
  class Input{
      BufferedReader br;
      StringTokenizer st;
      Input(){
          br=new BufferedReader(new InputStreamReader(System.in));
          st=new StringTokenizer("");
      }
      public int[] readArray() throws Exception{
          st=new StringTokenizer(br.readLine());
          int a[]=new int[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Integer.parseInt(st.nextToken());
          }
          return a;
      }
      public long[] readArrayLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          long a[]=new long[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Long.parseLong(st.nextToken());
          }
          return a;
      }
      public int readInt() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Integer.parseInt(st.nextToken());
      }
      public long readLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Long.parseLong(st.nextToken());
      }
      public String readString() throws Exception{
          return br.readLine();
      }
      public int[][] read2dArray(int n,int m)throws Exception{
          int a[][]=new int[n][m];
          for(int i=0;i<n;i++){
              st=new StringTokenizer(br.readLine());
              for(int j=0;j<m;j++){
                  a[i][j]=Integer.parseInt(st.nextToken());
              }
          }
          return a;
      }
  }
  class precalculates{
      public long gcd(long p, long q) {
          if (q == 0) return p;
          else return gcd(q, p % q);
      }
      public int[] prefixSumOneDimentional(int a[]){
          int n=a.length;
          int dp[]=new int[n];
          for(int i=0;i<n;i++){
              if(i==0)
                  dp[i]=a[i];
              else
                  dp[i]=dp[i-1]+a[i];
          }
          return dp;
      }
      public int[] postSumOneDimentional(int a[]) {
          int n = a.length;
          int dp[] = new int[n];
          for (int i = n - 1; i >= 0; i--) {
              if (i == n - 1)
                  dp[i] = a[i];
              else
                  dp[i] = dp[i + 1] + a[i];
          }
          return dp;
      }
      public int[][] prefixSum2d(int a[][]){
          int n=a.length;int m=a[0].length;
          int dp[][]=new int[n+1][m+1];
          for(int i=1;i<=n;i++){
              for(int j=1;j<=m;j++){
                  dp[i][j]=a[i-1][j-1]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
              }
          }
          return dp;
      }
      public long pow(long a,long b){
          long mod=1000000007;
          long ans=1;
          if(b<=0)
              return 1;
          if(b%2==0){
              ans=pow(a,b/2)%mod;
              return ((ans%mod)*(ans%mod))%mod;
          }else{
              ans=pow(a,b-1)%mod;
              return ((a%mod)*(ans%mod))%mod;
          }
      }
  
  }
  class GraphInteger{
      HashMap<Integer,vertex> vtces;
      class vertex{
          HashMap<Integer,Integer> children;
          public vertex(){
              children=new HashMap<>();
          }
      }
      public GraphInteger(){
          vtces=new HashMap<>();
      }
      public void addVertex(int a){
          vtces.put(a,new vertex());
      }
      public void addEdge(int a,int b,int cost){
          if(!vtces.containsKey(a)){
              vtces.put(a,new vertex());
          }
          if(!vtces.containsKey(b)){
              vtces.put(b,new vertex());
          }
          vtces.get(a).children.put(b,cost);
  //        vtces.get(b).children.put(a,cost);
      }
      public boolean isCyclicDirected(){
          boolean isdone[]=new boolean[vtces.size()+1];
          boolean check[]=new boolean[vtces.size()+1];
          for(int i=1;i<=vtces.size();i++) {
              if (!isdone[i] && isCyclicDirected(i,isdone, check)) {
                  return true;
              }
          }
          return false;
      }
      private boolean isCyclicDirected(int i,boolean isdone[],boolean check[]){
          if(check[i])
              return true;
          if(isdone[i])
              return false;
          check[i]=true;
          isdone[i]=true;
          Set<Integer> set=vtces.get(i).children.keySet();
          for(Integer ii:set){
              if(isCyclicDirected(ii,isdone,check))
                  return true;
          }
          check[i]=false;
          return false;
      }
  }
  class union_find {
      int n;
      int[] sz;
      int[] par;
  
      union_find(int nval) {
          n = nval;
          sz = new int[n + 1];
          par = new int[n + 1];
          for (int i = 0; i <= n; i++) {
              par[i] = i;
              sz[i] = 1;
          }
      }
  
      int root(int x) {
          if (par[x] == x)
              return x;
          return par[x] = root(par[x]);
      }
  
      boolean find(int a, int b) {
          return root(a) == root(b);
      }
  
      int union(int a, int b) {
          int ra = root(a);
          int rb = root(b);
          if (ra == rb)
              return 0;
          if(a==b)
              return 0;
          if (sz[a] > sz[b]) {
              int temp = ra;
              ra = rb;
              rb = temp;
          }
          par[ra] = rb;
          sz[rb] += sz[ra];
          return 1;
      }
  }
  
  
  
  
  
  
  
  