There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.

For example, suppose a=0111010000. 

  * In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00. 
  * In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100. 
  * It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1. 



Can you transform the string a into the string b using some finite number of operations (possibly, none)?

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.

The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.

The sum of n across all test cases does not exceed 3⋅ 10^5.

Output

For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).

Example

Input


5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100


Output


YES
YES
NO
YES
NO

Note

The first test case is shown in the statement.

In the second test case, we transform a into b by using zero operations.

In the third test case, there is no legal operation, so it is impossible to transform a into b.

In the fourth test case, here is one such transformation: 

  * Select the length 2 prefix to get 100101010101. 
  * Select the length 12 prefix to get 011010101010. 
  * Select the length 8 prefix to get 100101011010. 
  * Select the length 4 prefix to get 011001011010. 
  * Select the length 6 prefix to get 100110011010. 



In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class B {
    final int INF = 0x3f3f3f3f;
    BufferedReader  br;
    PrintWriter     out;
    StringTokenizer st;
    boolean         eof;

    public B() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) {
        new B();
    }

    private void solve() {
        int t = nextInt();
        while (t-- > 0) {
            doSolve();
        }
    }

    private void doSolve() {
        int n = nextInt();
        char[] from = nextString().toCharArray();
        char[] to = nextString().toCharArray();
        int[] flag = new int[from.length];
        int cnt0 = 0;
        int cnt1 = 0;
        for (int i = 0; i < from.length; i++) {
            if (from[i] == '0') ++cnt0;
            if (from[i] == '1') ++cnt1;
            if (cnt0 == cnt1) {
                flag[i] = 1;
            }
        }
        int cnt10 = 0;
        int cnt11 = 0;
        int[] flag1 = new int[to.length];
        for (int i = 0; i < to.length; i++) {
            if (to[i] == '0') ++cnt10;
            if (to[i] == '1') ++cnt11;
            if (cnt11 == cnt10) {
                flag1[i] = 1;
            }
        }


        for (int i = 0; i < flag.length; i++) {
            if (flag[i] != flag1[i]) {
                outln("NO");
                return;
            }
        }

        int i;
        for (i = flag.length - 1; i >= 0 && flag[i] != 1; --i) {
            if (from[i] != to[i]) {
                outln("NO");
                return;
            }
        }
        List<Integer> inx = new ArrayList<>();
        inx.add(-1);
        for (int j = 0; j < flag.length; ++j) {
            if (flag[j] == 1) {
                inx.add(j);
            }
        }
        for (int j = 0; j < inx.size() - 1; j++) {
            int start = inx.get(j);
            int end = inx.get(j + 1);
            Boolean reverse = null;
            for (int k = start+1; k <= end; ++k) {
                if (reverse == null) {
                    reverse = from[k] != to[k];
                } else {
                    if (reverse) {
                        if (from[k] == to[k]) {
                            outln("NO");
                            return;
                        }
                    } else {
                        if (from[k] != to[k]) {
                            outln("NO");
                            return;
                        }
                    }
                }
            }
        }

        outln("YES");
    }

    private void outln(Object o) {
        out.println(o);
    }

    private void out(Object o) {
        out.print(o);
    }

    public long[] nextLongArr(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public int[] nextIntArr(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() {
        return Double.parseDouble(nextToken());
    }

    class Point implements Comparable<Point> {
        int x;
        int y;

        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }


        @Override
        public String toString() {
            return "[" + x + "," + y + "]";
        }

        @Override
        public int compareTo(Point o) {
            if (this.x == o.x) {
                return this.y - o.y;
            }
            return this.x - o.x;
        }
    }


}