description: There are n candles on a Hanukkah menorah, and some of its candles are initially lit. We can describe which candles are lit with a binary string s, where the i-th candle is lit if and only if s_i=1.
  
  <image>
  
  Initially, the candle lights are described by a string a. In an operation, you select a candle that is currently lit. By doing so, the candle you selected will remain lit, and every other candle will change (if it was lit, it will become unlit and if it was unlit, it will become lit).
  
  You would like to make the candles look the same as string b. Your task is to determine if it is possible, and if it is, find the minimum number of operations required.
  
  Input
  
  The first line contains an integer t (1≤ t≤ 10^4) — the number of test cases. Then t cases follow.
  
  The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the number of candles.
  
  The second line contains a string a of length n consisting of symbols 0 and 1 — the initial pattern of lights.
  
  The third line contains a string b of length n consisting of symbols 0 and 1 — the desired pattern of lights.
  
  It is guaranteed that the sum of n does not exceed 10^5.
  
  Output
  
  For each test case, output the minimum number of operations required to transform a to b, or -1 if it's impossible.
  
  Example
  
  Input
  
  
  5
  5
  11010
  11010
  2
  01
  11
  3
  000
  101
  9
  100010111
  101101100
  9
  001011011
  011010101
  
  
  Output
  
  
  0
  1
  -1
  3
  4
  
  Note
  
  In the first test case, the two strings are already equal, so we don't have to perform any operations.
  
  In the second test case, we can perform a single operation selecting the second candle to transform 01 into 11.
  
  In the third test case, it's impossible to perform any operations because there are no lit candles to select.
  
  In the fourth test case, we can perform the following operations to transform a into b: 
  
    1. Select the 7-th candle: 100010{\color{red}1}11→ 011101{\color{red} 1}00. 
    2. Select the 2-nd candle: 0{\color{red} 1}1101100→ 1{\color{red} 1}0010011. 
    3. Select the 1-st candle: {\color{red}1}10010011→ {\color{red}1}01101100. 
  
  
  
  In the fifth test case, we can perform the following operations to transform a into b: 
  
    1. Select the 6-th candle: 00101{\color{red}1}011→ 11010{\color{red}1}100 
    2. Select the 2-nd candle: 1{\color{red}1}0101100→ 0{\color{red}1}1010011 
    3. Select the 8-th candle: 0110100{\color{red}1}1→ 1001011{\color{red}1}0 
    4. Select the 7-th candle: 100101{\color{red}1}10→ 011010{\color{red}1}01 
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.lang.*;
  import java.io.InputStreamReader;
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.File;
  import java.io.PrintStream;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  public class Main {
   
  	
  	/* 10^(7) = 1s.
  	 * ceilVal = (a+b-1) / b */
  	
  	static final int mod = 1000000007;
  	static final long temp = 998244353;
  	static final long MOD = 1000000007;
  	static final long M = (long)1e9+7;
   
  	
  	static class Pair implements Comparable<Pair> 
  	{
  		int first, second;
  		
  		public Pair(int first, int second) 
  		{
  			this.first = first;
  			this.second = second;
  		}
  		
  		public int compareTo(Pair ob) 
  		{
  			return (int)(first - ob.first);
  		}
  	}
  	
  	
  	static class Tuple implements Comparable<Tuple>
  	{
  		int first, second,third;
  		public Tuple(int first, int second, int third)
  		{
  			this.first = first;
  			this.second = second;
  			this.third = third;
  		}
  		public int compareTo(Tuple o)
  		{
  			return (int)(o.third - this.third);
  		}
  	}
  	
  	public static class DSU
  	{
  		int[] parent;
  		int[] rank; //Size of the trees is used as the rank
  		public DSU(int n)
  		{
  			parent = new int[n];
  			rank = new int[n];
  			Arrays.fill(parent, -1);
  			Arrays.fill(rank, 1);
  		}
  		
  		public int find(int i) //finding through path compression
  		{
  			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
  		}
  		
  		public boolean union(int a, int b) //Union Find by Rank 
  		{
  			a = find(a);
  			b = find(b);
  			
  			if(a == b) return false; //if they are already connected we exit by returning false.
  			
  			// if a's parent is less than b's parent
  			if(rank[a] < rank[b])
  			{
  				//then move a under b
  				parent[a] = b;
  			}
  			//else if rank of j's parent is less than i's parent
  			else if(rank[a] > rank[b])
  			{
  				//then move b under a
  				parent[b] = a;
  			}
  			//if both have the same rank.
  			else
  			{
  				//move a under b (it doesnt matter if its the other way around.
  				parent[b] = a;
  				rank[a] = 1 + rank[a];
  			}
  			return true;
  		}	
  	}
  
  	static class Reader 
  	{
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) throws IOException {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		
  		long[] longReadArray(int n) throws IOException {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  	
  	public static int gcd(int a, int b)
  	{
  		if(b == 0)
  		 return a;
  		
  		else
  		return gcd(b,a%b);
  	} 
   
  		public static long lcm(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  	public static long LongGCD(long a, long b)
  	{
  		if(b == 0)
  			 return a;
  			
  			else
  			return LongGCD(b,a%b);
  	}
  	
  		public static long LongLCM(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  		//Count the number of coprime's upto N
  		public static long phi(long n)  //euler totient/phi function
  	    { 
  	       long ans = n;
  //	       for(long i = 2;i*i<=n;i++)
  //	       {
  //	    	   if(n%i == 0)
  //	    	   {
  //	    		   while(n%i == 0) n/=i;
  //	    		   ans -= (ans/i);
  //	    	   }
  //	       }
  //	       
  //	       if(n > 1)
  //	       {
  //	    	   ans -= (ans/n);
  //	       }
  	       
  	       for(long i = 2;i<=n;i++)
  	       {
  	    	   if(isPrime(i))
  	    	   {
  	    		   ans -= (ans/i);
  	    	   }
  	       }
  	       
  	       return ans;
  	       
  	    }	
  	
  	public static long fastPow(long x, long n) 
  	{
  		if(n == 0)
  			return 1;
  		else if(n%2 == 0)
  			return fastPow(x*x,n/2);
  		else
  			return x*fastPow(x*x,(n-1)/2);
  	}
  	
  	   public static long modPow(long x, long y, long p)
  	    {
  	        long res = 1;
  	        x = x % p;
  	 
  	        while (y > 0) {
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	        return res;
  	    }
  	   
  	   static long modInverse(long n, long p)
  	    {
  	        return modPow(n, p - 2, p);
  	    }
  	 
  	    // Returns nCr % p using Fermat's little theorem.
  	    
  	    public static long nCrModP(long n, long r,long p)
  	    {
  	    	if (n<r) 
  	              return 0;
  	          
  	        if (r == 0)
  	            return 1;
  	 
  	        long[] fac = new long[(int)(n) + 1];
  	        fac[0] = 1;
  	 
  	        for (int i = 1; i <= n; i++)
  	            fac[i] = fac[i - 1] * i % p;
  	 
  	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
  	                % p * modInverse(fac[(int)(n - r)], p)
  	                % p)
  	            % p;
  	    }
  	    
  	    public static long fact(long n) 
  	    {
  	    	  long[] fac = new long[(int)(n) + 1];
  		        fac[0] = 1;
  		 
  		        for (long i = 1; i <= n; i++)
  		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
  		        
  		        return fac[(int)(n)];
  		}
  	    
  	    public static long nCr(long n, long k) 
  	    {
  			long ans = 1;
  			for(long i = 0;i<k;i++)
  			{
  				ans *= (n-i);
  				ans /= (i+1);
  			}
  			return ans;
  		}
   
  		//Modular Operations for Addition and Multiplication. 
  		   public static long perfomMod(long x)
  		   {
  		        return ((x%M + M)%M);
  		    }
  		   public static long addMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)+perfomMod(b));
  		    }
  		  
  		   public static long subMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)-perfomMod(b));
  		   }
  		   
  		   public static long mulMod(long  a, long b)
  		   {
  		        return perfomMod(perfomMod(a)*perfomMod(b));
  		   }
  		   
  		   public static boolean isPrime(long n) 
  			{
  				if(n == 1)
  				{
  					return false;
  				}
  				
  				//check only for sqrt of the number as the divisors 
  				//keep repeating so only half of them are required. So,sqrt.
  				for(int i = 2;i*i<=n;i++)
  				{
  					if(n%i == 0)
  					{
  						return false;
  					}
  				}
  				return true;
  			}
  			
  			public static List<Integer> SieveList(int n)
  			{
  				boolean prime[] = new boolean[(int)(n+1)];
  				Arrays.fill(prime, true);
  				
  				List<Integer> l = new ArrayList<>();
  				for (int p = 2; p*p<=n; p++) 
  				{ 
  					if (prime[p] == true) 
  					{ 	
  					    for(int i = p*p; i<=n; i += p) 
  						{
  						    prime[i] = false; 
  						}	
  					} 
  				} 
  		 
  				for (int p = 2; p<=n; p++) 
  				{
  				    if (prime[p] == true)
  				    {
  				       l.add(p); 
  				    }
  				}
  				
  				return l;
  			}
  			
  			public static int countDivisors(int x)
  			{
  				int c = 0;
  				for(int i = 1;i*i<=x;i++)
  				{
  					if(x%i == 0)
  					{
  						if(x/i != i)
  						{
  							c+=2;
  						}
  						else 
  						{
  							c++;
  						}
  					}
  				}
  				return c;
  			}
  			
  			public static long log2(long n)
  		    {
  				 long ans = (long)(log(n)/log(2));
  				 return ans;
  			}
  			   
  			public static boolean isPow2(long n)
  			{
  				 return (n  != 0 && ((n & (n-1))) == 0);
  			}	
  		   
  		   public static boolean isSq(int x)  
  		   {
  			    long s = (long)Math.round(Math.sqrt(x));
  			    return s*s==x;
  		   }
  		
  			   /*
  			    * 
  			    * >=         <=
  			   0 1 2 3 4 5 6 7  
  			   5 5 5 6 6 6 7 7
  			   lower_bound for 6 at index 3 (>=)
  			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
  			   */
  
  		   public static int LowerBound(int a[], int x) 
  		   {
  			      int l=-1,r=a.length;
  				  while(l+1<r) 
  				  {
  				    int m=(l+r)>>>1;
  				    if(a[m]>=x) r=m;
  				    else l=m;
  				  }
  				  return r;
  			}
  		   
  			public static int UpperBound(long a[], long x) 
  			{
  				int l=-1, r=a.length;
  			    while(l+1<r) 
  			    {
  			       int m=(l+r)>>>1;
  			       if(a[m]<=x) l=m;
  			       else r=m;
  			    }
  			    return l+1;
  			}
  			
  			public static void Sort(long[] a) 
  			{
  				List<Long> l = new ArrayList<>();
  				for (long i : a) l.add(i);
  				Collections.sort(l);
  //				Collections.reverse(l);  //Use to Sort decreasingly 
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  			
  			public static void ssort(char[] a)
  			{
  				List<Character> l = new ArrayList<>();
  				for (char i : a) l.add(i);
  				Collections.sort(l);
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  	
  	
  	public static void main(String[] args) throws Exception 
  	{
  		Reader sc = new Reader();
  		PrintWriter fout = new PrintWriter(System.out); 
  		
  		int tt = sc.nextInt();
  		while(tt-- > 0)
  		{
  	       int n = sc.nextInt();
  	       char[] a = sc.next().toCharArray(), b = sc.next().toCharArray();
  	       
  	       int c00 = 0, c01 = 0, c10 = 0, c11 = 0;
  	       for(int i = 0;i<n;i++)
  	       {
  	    	   if(a[i] == '0' && b[i] == '0')
  	    	   {
  	    		   c00++;
  	    	   }
  	    	   else if(a[i] == '0' && b[i] == '1')
  	    	   {
  	    		   c01++;
  	    	   }
  	    	   else if(a[i] == '1' && b[i] == '0')
  	    	   {
  	    		   c10++;
  	    	   }
  	    	   else if(a[i] == '1' && b[i] == '1')
  	    	   {
  	    		   c11++;
  	    	   }
  	       }
  	       
  	       int ans = mod;
  	       if(c01 == c10) ans = min(ans, c01 + c10);
  	       if(c11 == c00 + 1) ans = min(ans, c11 + c00);
  	       
  	       fout.println((ans == mod) ? -1 : ans);
  		}
  		
  		fout.close();
  	}
  }