Little Dima has two sequences of points with integer coordinates: sequence (a1, 1), (a2, 2), ..., (an, n) and sequence (b1, 1), (b2, 2), ..., (bn, n).

Now Dima wants to count the number of distinct sequences of points of length 2·n that can be assembled from these sequences, such that the x-coordinates of points in the assembled sequence will not decrease. Help him with that. Note that each element of the initial sequences should be used exactly once in the assembled sequence.

Dima considers two assembled sequences (p1, q1), (p2, q2), ..., (p2·n, q2·n) and (x1, y1), (x2, y2), ..., (x2·n, y2·n) distinct, if there is such i (1 ≤ i ≤ 2·n), that (pi, qi) ≠ (xi, yi).

As the answer can be rather large, print the remainder from dividing the answer by number m.

Input

The first line contains integer n (1 ≤ n ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). The third line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 109). The numbers in the lines are separated by spaces.

The last line contains integer m (2 ≤ m ≤ 109 + 7).

Output

In the single line print the remainder after dividing the answer to the problem by number m. 

Examples

Input

1
1
2
7


Output

1


Input

2
1 2
2 3
11


Output

2

Note

In the first sample you can get only one sequence: (1, 1), (2, 1). 

In the second sample you can get such sequences : (1, 1), (2, 2), (2, 1), (3, 2); (1, 1), (2, 1), (2, 2), (3, 2). Thus, the answer is 2.
import java.io.*;
import java.util.*;

public class A implements Runnable {
	public static void main (String[] args) {new Thread(null, new A(), "_cf", 1 << 28).start();}
	
	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		System.err.println("Go!");

		int n = fs.nextInt();
		int[] a = fs.nextIntArray(n);
		int[] b = fs.nextIntArray(n);
		int mod = fs.nextInt();
		Pair[] ps = new Pair[2 * n];
		for(int i = 0; i < ps.length; i++) {
			if(i < n) ps[i] = new Pair(a[i], i);
			else ps[i] = new Pair(b[i-n], i-n);
		}
		Arrays.sort(ps);
		long res = 1;
		for(int i = 0; i < ps.length; i++) {
			int count = 0;
			int j = i;
			while(j < ps.length && ps[j].value == ps[i].value) j++;
			count = j-i;
			int facts = 0;
			for(int ii = i; ii < j-1; ii++) {
				if(ps[ii].index == ps[ii+1].index) facts++;
			}
			long mult = factorial(count, facts, mod);
			res = (res * mult) % mod;
			i = --j;
		}
		out.println(res);
		out.close();
	}
	
	int factorial(int n, int twos, int mod) {
		long res = 1;
		for(int i = 1; i <= n; i++) {
			long mult = i;
			while(twos > 0 && mult % 2 == 0) {
				mult /= 2;
				twos--;
			}
			res = (res * mult) % mod;
		}
		
		return (int)res;
	}
	
	class Pair implements Comparable<Pair> {
		int value, index;
		Pair(int a, int b) {
			value = a; index = b;
		}
		public int compareTo(Pair p) {
			int comp = Integer.compare(value, p.value);
			if(comp == 0) comp = Integer.compare(index, p.index);
			return comp;
		}
		public String toString() {
			return String.format("(%d, %d)", value, index);
		}
	}
	
	class FastScanner {
		public int BS = 1<<16;
		public char NC = (char)0;
		byte[] buf = new byte[BS];
		int bId = 0, size = 0;
		char c = NC;
		double num = 1;
		BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) throws FileNotFoundException {
			in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
		}

		public char nextChar(){
			while(bId==size) {
				try {
					size = in.read(buf);
				}catch(Exception e) {
					return NC;
				}                
				if(size==-1)return NC;
				bId=0;
			}
			return (char)buf[bId++];
		}

		public int nextInt() {
			return (int)nextLong();
		}

		public long nextLong() {
			num=1;
			boolean neg = false;
			if(c==NC)c=nextChar();
			for(;(c<'0' || c>'9'); c = nextChar()) {
				if(c=='-')neg=true;
			}
			long res = 0;
			for(; c>='0' && c <='9'; c=nextChar()) {
				res = (res<<3)+(res<<1)+c-'0';
				num*=10;
			}
			return neg?-res:res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c!='.' ? cur:cur+nextLong()/num;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c>32) {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c!='\n') {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if(c>32)return true;
			while(true) {
				c=nextChar();
				if(c==NC)return false;
				else if(c>32)return true;
			}
		}
		
		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for(int i = 0; i < n; i++) res[i] = nextInt();
			return res;
		}
		
	}

	
}