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.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class cf1506f {

    public static void main(String[] args) throws IOException {
        int t = ri();
        while (t --> 0) {
            int n = ri(), p[][] = new int[n][2];
            r();
            for (int i = 0; i < n; ++i) {
                p[i][0] = ni();
            }
            r();
            for (int i = 0; i < n; ++i) {
                p[i][1] = ni();
            }
            sort(p, (a, b) -> a[0] - b[0]);
            int ch[] = new int[n];
            for (int i = 0; i < n; ++i) {
                ch[i] = (p[i][0] - p[i][1]) / 2;
            }
            int cur = 0, cur_row = 1;
            long ans = 0;
            for (int i = 0; i < n; ++i) {
                if ((p[i][0] + p[i][1]) % 2 == 0 && ch[i] == cur) {
                    ans += p[i][0] - cur_row;
                } else {
                    ans += abs(cur - ch[i]);
                }
                cur = ch[i];
                cur_row = p[i][0];
            }
            prln(ans);
        }
        close();
    }

    static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;
    static Random __r = new Random();

    // references
    // IBIG = 1e9 + 7
    // IMAX ~= 2e9
    // LMAX ~= 9e18

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final long LMAX = 9223372036854775807L;
    // math util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int fli(double d) {return (int) d;}
    static int cei(double d) {return (int) ceil(d);}
    static long fll(double d) {return (long) d;}
    static long cel(double d) {return (long) ceil(d);}
    static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
    static long gcd(long a, long b) {return b == 0 ? a : gcd(b, a % b);}
    static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
    static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
    static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
    static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
    // array util
    static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void rsort(int[] a) {shuffle(a); sort(a);}
    static void rsort(long[] a) {shuffle(a); sort(a);}
    static void rsort(double[] a) {shuffle(a); sort(a);}
    static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    // input
    static void r() throws IOException {input = new StringTokenizer(rline());}
    static int ri() throws IOException {return Integer.parseInt(rline());}
    static long rl() throws IOException {return Long.parseLong(rline());}
    static double rd() throws IOException {return Double.parseDouble(rline());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
    static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
    static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
    static char[] rcha() throws IOException {return rline().toCharArray();}
    static String rline() throws IOException {return __i.readLine();}
    static String n() {return input.nextToken();}
    static int rni() throws IOException {r(); return ni();}
    static int ni() {return Integer.parseInt(n());}
    static long rnl() throws IOException {r(); return nl();}
    static long nl() {return Long.parseLong(n());}
    static double rnd() throws IOException {r(); return nd();}
    static double nd() {return Double.parseDouble(n());}
    // output
    static void pr(int i) {__o.print(i);}
    static void prln(int i) {__o.println(i);}
    static void pr(long l) {__o.print(l);}
    static void prln(long l) {__o.println(l);}
    static void pr(double d) {__o.print(d);}
    static void prln(double d) {__o.println(d);}
    static void pr(char c) {__o.print(c);}
    static void prln(char c) {__o.println(c);}
    static void pr(char[] s) {__o.print(new String(s));}
    static void prln(char[] s) {__o.println(new String(s));}
    static void pr(String s) {__o.print(s);}
    static void prln(String s) {__o.println(s);}
    static void pr(Object o) {__o.print(o);}
    static void prln(Object o) {__o.println(o);}
    static void prln() {__o.println();}
    static void pryes() {prln("yes");}
    static void pry() {prln("Yes");}
    static void prY() {prln("YES");}
    static void prno() {prln("no");}
    static void prn() {prln("No");}
    static void prN() {prln("NO");}
    static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
    static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
    static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
    static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
    static void h() {prln("hlfd"); flush();}
    static void flush() {__o.flush();}
    static void close() {__o.close();}
}