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.*;
import java.util.Objects;
import java.util.StringTokenizer;

import static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.exit;


public class Solution {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;
    static boolean isLocal = false;

    void Case() throws IOException {
        int n = nextInt(), bal = 0;
        char[] a = next().toCharArray();
        char[] b = next().toCharArray();
        int dp[] = new int[n];
        for (int i = 0; i < n; i++) {
            if (i == 0) dp[i] = (a[i] == '1' ? 1 : -1);
            else dp[i] = dp[i - 1] + (a[i] == '1' ? 1 : -1);
        }
        boolean reversing = false;
        for (int i = n - 1; i >= 0; i--) {
            if (a[i] != b[i]) {
                if (reversing) continue;
                if (dp[i] == 0) {
                    reversing = true;
                } else if (dp[i] != 0) {
                    out.println("NO");
                    return;
                }
            } else {
                if (reversing) {
                    if (dp[i] != 0) {
                        out.println("NO");
                        return;
                    } else reversing = false;
                }
            }
        }
        out.println("YES");
    }

    void solve() throws Exception {
        int t = nextInt();
        for (int i = 1; i <= t; i++) {
//            out.println("Case #" + i + ":");
            Case();
        }
    }

    private int gcd(int a, int b) {
        while (b > 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    int min(int x, int y) {
        return Integer.min(x, y);
    }

    int max(int x, int y) {
        return Integer.max(x, y);
    }

    long min(long x, long y) {
        return Long.min(x, y);
    }

    long max(long x, long y) {
        return Long.max(x, y);
    }

    int[] sort(int[] arr) {
        sort(arr, 0, arr.length - 1);
        return arr;
    }

    void sort(int arr[], int l, int r) {
        if (l < r) {
            int m = (l + r) / 2;
            sort(arr, l, m);
            sort(arr, m + 1, r);
            merge(arr, l, m, r);
        }
    }

    void merge(int arr[], int l, int m, int r) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int L[] = new int[n1];
        int R[] = new int[n2];
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];
        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    class Seg implements Comparable<Seg> {
        int st, end;

        public Seg(int st, int end) {
            this.st = st;
            this.end = end;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Seg seg = (Seg) o;
            return st == seg.st &&
                    end == seg.end;
        }

        @Override
        public int hashCode() {
            return Objects.hash(st, end);
        }

        @Override
        public int compareTo(Seg seg) {
            return st == seg.st ? Integer.compare(end, seg.end) : Integer.compare(st, seg.st);
        }
    }

    private int[] na(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    private long[] nal(int n) throws IOException {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = nextLong();
        return a;
    }

    int nextInt() throws IOException {
        return parseInt(next());
    }

    long nextLong() throws IOException {
        return parseLong(next());
    }

    double nextDouble() throws IOException {
        return parseDouble(next());
    }

    String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) throws Exception {
        try {
            if (isLocal) {
                in = new BufferedReader(new FileReader("src/tests/sol.in"));
                out = new PrintWriter(new BufferedWriter(new FileWriter("src/tests/sol.out")));
            } else {
                in = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(new OutputStreamWriter(System.out));
            }
//            long lStartTime = System.currentTimeMillis();
            new Solution().solve();
//            long lEndTime = System.currentTimeMillis();
//            out.println("Elapsed time in seconds: " + (double) (lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}
