Consider an infinite triangle made up of layers. Let's number the layers, starting from one, from the top of the triangle (from top to bottom). The k-th layer of the triangle contains k points, numbered from left to right. Each point of an infinite triangle is described by a pair of numbers (r, c) (1 ≤ c ≤ r), where r is the number of the layer, and c is the number of the point in the layer. From each point (r, c) there are two directed edges to the points (r+1, c) and (r+1, c+1), but only one of the edges is activated. If r + c is even, then the edge to the point (r+1, c) is activated, otherwise the edge to the point (r+1, c+1) is activated. Look at the picture for a better understanding.

<image> Activated edges are colored in black. Non-activated edges are colored in gray.

From the point (r_1, c_1) it is possible to reach the point (r_2, c_2), if there is a path between them only from activated edges. For example, in the picture above, there is a path from (1, 1) to (3, 2), but there is no path from (2, 1) to (1, 1).

Initially, you are at the point (1, 1). For each turn, you can: 

  * Replace activated edge for point (r, c). That is if the edge to the point (r+1, c) is activated, then instead of it, the edge to the point (r+1, c+1) becomes activated, otherwise if the edge to the point (r+1, c+1), then instead if it, the edge to the point (r+1, c) becomes activated. This action increases the cost of the path by 1; 
  * Move from the current point to another by following the activated edge. This action does not increase the cost of the path. 



You are given a sequence of n points of an infinite triangle (r_1, c_1), (r_2, c_2), …, (r_n, c_n). Find the minimum cost path from (1, 1), passing through all n points in arbitrary order.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) is the number of test cases. Then t test cases follow.

Each test case begins with a line containing one integer n (1 ≤ n ≤ 2 ⋅ 10^5) is the number of points to visit.

The second line contains n numbers r_1, r_2, …, r_n (1 ≤ r_i ≤ 10^9), where r_i is the number of the layer in which i-th point is located.

The third line contains n numbers c_1, c_2, …, c_n (1 ≤ c_i ≤ r_i), where c_i is the number of the i-th point in the r_i layer.

It is guaranteed that all n points are distinct.

It is guaranteed that there is always at least one way to traverse all n points.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output the minimum cost of a path passing through all points in the corresponding test case.

Example

Input


4
3
1 4 2
1 3 1
2
2 4
2 3
2
1 1000000000
1 1000000000
4
3 10 5 8
2 5 2 4


Output


0
1
999999999
2
import java.util.*;
import java.io.*;
import java.math.*;

/**
 *
 * @Har_Har_Mahadev
 */

public class F {

	private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
	private static int N = 0;

	public static void process() throws IOException {

		int n = sc.nextInt();
		ArrayList<Pair> lis = new ArrayList<F.Pair>();
		lis.add(new Pair(1, 1));
		long a[] = sc.readArrayLong(n);
		long b[] = sc.readArrayLong(n);
		
		for(int i = 0; i<n; i++) {
			lis.add(new Pair(a[i], b[i]));
		}
		Collections.sort(lis);
		long ans = 0;
		for(int i = 0; i<n; i++) {
			ans+=solve(lis.get(i).x,lis.get(i).y,lis.get(i+1).x,lis.get(i+1).y);
		}
		System.out.println(ans);

	}

	private static long solve(long fromX, long fromY, long toX, long toY) {
		if (fromX==toX) return 0;
	    long dR=toX-fromX;
	    if ((fromX+fromY)%2!=0) {
	        //adding is free, going left twice costs 1
	    	long targetC=fromY+dR;
	        return (targetC-toY+1)/2;
	    }
	    else {
	    	long targetC=toY-fromY;
	        //if we need to go right the entire time, do that now
	        if (targetC==dR) return dR;
	        //otherwise go left for free here
	        return solve(fromX+1, fromY, toX, toY);
	    }
	}

	//=============================================================================
	//--------------------------The End---------------------------------
	//=============================================================================

	static FastScanner sc;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		boolean oj = true;
		if (oj) {
			sc = new FastScanner();
			out = new PrintWriter(System.out);
		} else {
			sc = new FastScanner(100);
			out = new PrintWriter("output.txt");
		}
		int t = 1;
		t = sc.nextInt();
		while (t-- > 0) {
			process();
		}
		out.flush();
		out.close();
	}

	static class Pair implements Comparable<Pair> {
		long x, y;

		Pair(long x, long y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Pair o) {
			return Long.compare(this.x, o.x);
		}

		//		 @Override
		//		    public boolean equals(Object o) {
		//		        if (this == o) return true;
		//		        if (!(o instanceof Pair)) return false;
		//		        Pair key = (Pair) o;
		//		        return x == key.x && y == key.y;
		//		    }
		//		 
		//		    @Override
		//		    public int hashCode() {
		//		        int result = x;
		//		        result = 31 * result + y;
		//		        return result;
		//		    }
	}

	/////////////////////////////////////////////////////////////////////////////////////////////////////////

	static void println(Object o) {
		out.println(o);
	}

	static void println() {
		out.println();
	}

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

	static void pflush(Object o) {
		out.println(o);
		out.flush();
	}

	static int ceil(int x, int y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static long ceil(long x, long y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

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

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

	static int abs(int x) {
		return Math.abs(x);
	}

	static long abs(long x) {
		return Math.abs(x);
	}

	static int log2(int N) {
		int result = (int) (Math.log(N) / Math.log(2));
		return result;
	}

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

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

	public static int gcd(int a, int b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.intValue();
	}

	public static long gcd(long a, long b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.longValue();
	}

	public static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static int lcm(int a, int b) {
		return (a * b) / gcd(a, b);
	}

	public static int lower_bound(int[] arr, int x) {
		int low = 0, high = arr.length, mid = -1;

		while (low < high) {
			mid = (low + high) / 2;

			if (arr[mid] >= x)
				high = mid;
			else
				low = mid + 1;
		}

		return low;
	}

	public static int upper_bound(int[] arr, int x) {
		int low = 0, high = arr.length, mid = -1;

		while (low < high) {
			mid = (low + high) / 2;

			if (arr[mid] > x)
				high = mid;
			else
				low = mid + 1;
		}

		return low;
	}

	/////////////////////////////////////////////////////////////////////////////////////////////////////////

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner() throws FileNotFoundException {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		FastScanner(int a) throws FileNotFoundException {
			br = new BufferedReader(new FileReader("input.txt"));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

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

		String nextLine() throws IOException {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] readArray(int n) throws IOException {
			int[] A = new int[n];
			for (int i = 0; i != n; i++) {
				A[i] = sc.nextInt();
			}
			return A;
		}

		long[] readArrayLong(int n) throws IOException {
			long[] A = new long[n];
			for (int i = 0; i != n; i++) {
				A[i] = sc.nextLong();
			}
			return A;
		}
	}

	static void ruffleSort(int[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}

	static void ruffleSort(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
		Arrays.sort(a);
	}
}
