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.

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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 
 //package eround101;

import java.util.*;
import java.io.*;
import java.lang.*;
import java.util.StringTokenizer;

public class Solution {

    static HritikScanner sc = new HritikScanner();
    static PrintWriter pw = new PrintWriter(System.out, true);
    final static int MOD = 1000000007;

    public static void main(String[] args) {

        int t = ni();
        
        while (t-- > 0) {
            solve();
        }
    }

    static void solve() {
        int n = ni();
        char[] a = sc.next().toCharArray();
        char[] b = sc.next().toCharArray();
        int same = 0, same1 = 0, same0 = 0;
        int diff = 0, diff1 = 0, diff0 = 0;
        for(int i = 0; i < n; i++)
        {
            if(a[i] == b[i])
            {
                same++;
                if(a[i] == '1')
                    same1++;
                else
                    same0++;
            }
            else
            {
                diff++;
                if(a[i] == '0')
                    diff0++;
                else
                    diff1++;
            }
        }
        int ans = Integer.MAX_VALUE;
        if(diff % 2 == 0 && diff1 == diff0)
        {
            ans = Math.min(ans, diff);
        }
        if(same % 2 == 1 && same1 == same0+1)
        {
            ans = Math.min(ans, same);
        }
        pl(ans == Integer.MAX_VALUE ? -1 : ans);
    }
    /////////////////////////////////////////////////////////////////////////////////

    static class FenwickTree { // Binary Index Tree

        int[] tree;
        static int size;

        public FenwickTree(int size) {
            this.size = size;
            tree = new int[size + 5];
        }

        public void add(int i, int val) {
            while (i <= size) {
                tree[i] += val;
                i += i & -i;  // adding the decimal value of the last set bit.
            }
        }

        public int sum(int i) {
            int res = 0;
            while (i >= 1) {
                res += tree[i];
                i -= i & -i; // deleting the last set bit
            }
            return res;
        }

        public int sum(int l, int r) {
            return sum(r) - sum(l - 1);
        }
    }

    /////////////////////////////////////////////////////////////////////////////////
    static int[] nextIntArray(int n) {
        int[] arr = new int[n];
        int i = 0;
        while (i < n) {
            arr[i++] = ni();
        }
        return arr;
    }

    static long[] nextLongArray(int n) {
        long[] arr = new long[n];
        int i = 0;
        while (i < n) {
            arr[i++] = nl();
        }
        return arr;
    }

    static int[] nextIntArray1(int n) {
        int[] arr = new int[n + 1];
        int i = 1;
        while (i <= n) {
            arr[i++] = ni();
        }
        return arr;
    }

    /////////////////////////////////////////////////////////////////////////////////
    static int ni() {
        return sc.nextInt();
    }

    static long nl() {
        return sc.nextLong();
    }

    static double nd() {
        return sc.nextDouble();
    }

    /////////////////////////////////////////////////////////////////////////////////
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o + " ");
    }

    static void pl(Object o) {
        pw.println(o);
    }

/////////////////////////////////////////////////////////////////////////////////
    static void print(Object s) {
        System.out.println(s);
    }

    /////////////////////////////////////////////////////////////////////////////////
    //Arrays.sort(arr(type int)) uses quick sort whose worst case time complexity is O(N^2) in case of sorted array.
    // we can either use Object (Integer, Long)[Here it will use merge sort] array or suffle the array before sorting
    // or we can use Collections.sort(list))
    public static long[] sort(long arr[]){
            List<Long> list = new ArrayList<>();
            for(long n : arr){list.add(n);}
            Collections.sort(list);
            for(int i=0;i<arr.length;i++){
                arr[i] = list.get(i);
            }
            return arr;
        }
 
        public static int[] sort(int arr[]){
            List<Integer> list = new ArrayList<>();
            for(int n : arr){list.add(n);}
            Collections.sort(list);
            for(int i=0;i<arr.length;i++){
                arr[i] = list.get(i);
            }
            return arr;
        }
    /////////////////////////////////////////////////////////////////////////////////
    //-----------HritikScanner class for faster input----------//
    static class HritikScanner {

        BufferedReader br;
        StringTokenizer st;

        public HritikScanner() {
            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;
        }
    }

    //////////////////////////////////////////////////////////////////
    public static class Pair implements Comparable<Pair> {

        int num, row, col;

        Pair(int num, int row, int col) {
            this.num = num;
            this.row = row;
            this.col = col;
        }

        public int get1() {
            return num;
        }

        public int get2() {
            return row;
        }

        public int compareTo(Pair A) {
            return this.num - A.num;
        }

        public String toString() {
            return num + " " + row + " " + col;
        }
    }

    //////////////////////////////////////////////////////////////////
    // Function to return gcd of a and b time complexity O(log(a+b))
    static long gcd(long a, long b) {
        if (a == 0) {
            return b;
        }
        return gcd(b % a, a);
    }

    // method to return LCM of two numbers 
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    //////////////////////////////////////////////////////////////////
    static boolean isPrime(long n) {
        // Corner cases 
        if (n <= 1) {
            return false;
        }
        if (n <= 3) {
            return true;
        }

        // This is checked so that we can skip  
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) {
            return false;
        }

        for (long i = 5; i * i <= n; i = i + 6) {
            if (n % i == 0 || n % (i + 2) == 0) {
                return false;
            }
        }

        return true;
    }
    //////////////////////////////////////////////////////////////////

    static boolean isPowerOfTwo(long n) {
        if (n == 0) {
            return false;
        }

        return (long) (Math.ceil((Math.log(n) / Math.log(2))))
                == (long) (Math.floor(((Math.log(n) / Math.log(2)))));
    }

    public static long getFact(int n) {
        long ans = 1;
        while (n > 0) {
            ans *= n;
            ans %= MOD;
            n--;
        }
        return ans;
    }

    public static long pow(long n, int pow) {
        if (pow == 0) {
            return 1;
        }
        long temp = pow(n, pow / 2) % MOD;
        temp *= temp;
        temp %= MOD;
        if (pow % 2 == 1) {
            temp *= n;
        }
        temp %= MOD;
        return temp;

    }

    public static long nCr(int n, int r) {
        long ans = 1;
        int temp = n - r;
        while (n > temp) {
            ans *= n;
            ans %= MOD;
            n--;
        }
        ans *= pow(getFact(r) % MOD, MOD - 2) % MOD;
        ans %= MOD;
        return ans;
    }

    //////////////////////////////////////////////////////////////////
    // method returns Nth power of A 
    static double nthRoot(int A, int N) {

        // intially guessing a random number between 
        // 0 and 9 
        double xPre = Math.random() % 10;

        // smaller eps, denotes more accuracy 
        double eps = 0.001;

        // initializing difference between two 
        // roots by INT_MAX 
        double delX = 2147483647;

        // xK denotes current value of x 
        double xK = 0.0;

        // loop untill we reach desired accuracy 
        while (delX > eps) {
            // calculating current value from previous 
            // value by newton's method 
            xK = ((N - 1.0) * xPre
                    + (double) A / Math.pow(xPre, N - 1)) / (double) N;
            delX = Math.abs(xK - xPre);
            xPre = xK;
        }

        return xK;
    }

}
