Monocarp had a sequence a consisting of n + m integers a_1, a_2, ..., a_{n + m}. He painted the elements into two colors, red and blue; n elements were painted red, all other m elements were painted blue.

After painting the elements, he has written two sequences r_1, r_2, ..., r_n and b_1, b_2, ..., b_m. The sequence r consisted of all red elements of a in the order they appeared in a; similarly, the sequence b consisted of all blue elements of a in the order they appeared in a as well.

Unfortunately, the original sequence was lost, and Monocarp only has the sequences r and b. He wants to restore the original sequence. In case there are multiple ways to restore it, he wants to choose a way to restore that maximizes the value of 

$$$f(a) = max(0, a_1, (a_1 + a_2), (a_1 + a_2 + a_3), ..., (a_1 + a_2 + a_3 + ... + a_{n + m}))$$$

Help Monocarp to calculate the maximum possible value of f(a).

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then the test cases follow. Each test case consists of four lines.

The first line of each test case contains one integer n (1 ≤ n ≤ 100).

The second line contains n integers r_1, r_2, ..., r_n (-100 ≤ r_i ≤ 100).

The third line contains one integer m (1 ≤ m ≤ 100).

The fourth line contains m integers b_1, b_2, ..., b_m (-100 ≤ b_i ≤ 100).

Output

For each test case, print one integer — the maximum possible value of f(a).

Example

Input


4
4
6 -5 7 -3
3
2 3 -4
2
1 1
4
10 -3 2 2
5
-1 -2 -3 -4 -5
5
-1 -2 -3 -4 -5
1
0
1
0


Output


13
13
0
0

Note

In the explanations for the sample test cases, red elements are marked as bold.

In the first test case, one of the possible sequences a is [6, 2, -5, 3, 7, -3, -4].

In the second test case, one of the possible sequences a is [10, 1, -3, 1, 2, 2].

In the third test case, one of the possible sequences a is [-1, -1, -2, -3, -2, -4, -5, -3, -4, -5].

In the fourth test case, one of the possible sequences a is [0, 0].
import java.io.DataInputStream;
import java.io.IOException;
import java.util.List;

public class A {

	
	public static void pr(int i) {
		System.out.println(i);
	}
	
	public static int max(int[] ar) {
		int max = Integer.MIN_VALUE;
		for (int i : ar) {
			max = Math.max(max, i);
		}
		return max;
	}

	public static long max(long[] ar) {
		long max = Long.MIN_VALUE;
		for (long i : ar) {
			max = Math.max(max, i);
		}
		return max;
	}

	public static long max(List<Long> ar) {
		long max = Long.MIN_VALUE;
		for (long i : ar) {
			max = Math.max(max, i);
		}
		return max;
	}

	static int testCase(int[] ar, int r, int[] ab, int b) {
		int maxr = ar[0], maxb = ab[0];
		for(int i = 1; i < r; i++) {
			ar[i] += ar[i-1];
			maxr = Math.max(maxr, ar[i]);
		}
		
		for(int i = 1; i < b; i++) {
			ab[i] += ab[i-1];
			maxb = Math.max(maxb, ab[i]);
		}
		return Math.max(maxr, 0) + Math.max(maxb, 0);
	}
	
	public static void main(String[] args) throws IOException {
		Reader sc = new Reader();
		int t = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		while (t-- > 0) { //Longest Common Substring
			int r;
			r = sc.nextInt();
			int[] ar = new int[r];
			
			for(int i = 0; i < r; i++) {
				ar[i] = sc.nextInt();
			}
			int b;
			b = sc.nextInt();
			int[] ab = new int[b];
			for(int i = 0; i < b; i++) {
				ab[i] = sc.nextInt();
			}
			sb.append(testCase(ar, r, ab, b)).append("\n");
		}
		System.out.println(sb);
	}

	protected static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;
		private final int al = 1024;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String next() throws IOException {
			byte[] buf = new byte[al];
			int cnt = 0, c = 0;
			while ((c = read()) != -1) {
				if (c == ' ')
					break;
				if (c == '\n') {
					cnt--;
					break;
				}
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt == -1 ? 0 : cnt);
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[al]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n') {
					cnt--;
					break;
				}
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt == -1 ? 0 : cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

}
