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.
/*
Keep solving problems.
*/

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

public class CFA {
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    final long MOD = 1000L * 1000L * 1000L + 7;
    int[] dx = {0, -1, 0, 1};
    int[] dy = {1, 0, -1, 0};
    private static final String yes = "YES";
    private static final String no = "NO";

    void solve() throws IOException {
        int n = nextInt();
        int[] a = nextIntArr(n);
        int[] b = nextIntArr(n);
        long m = nextLong();
        int len = 200000 + 10;
        int[] cnt2 = new int[len];
        for (int i = 2; i < len; i++) {
            int cnt = 0;
            int j = i;
            while (j % 2 == 0) {
                cnt++;
                j = j / 2;
            }
            cnt2[i] = cnt;
        }
        int[] sum = new int[len];
        for (int i = 1; i < len; i++) {
            sum[i] = sum[i - 1] + cnt2[i];
        }
        long[] fact = new long[len];
        fact[0] = 1;
        fact[1] = 1;
        for (int i = 2; i < len; i++) {
            int j = i;
            while (j % 2 == 0) {
                j = j / 2;
            }
            fact[i] = fact[i - 1] * j;
            fact[i] %= m;
        }

        Map<Integer, Integer> hm = new HashMap<>();
        for (int v : a) {
            hm.put(v, hm.getOrDefault(v, 0) + 1);
        }
        for (int v : b) {
            hm.put(v, hm.getOrDefault(v, 0) + 1);
        }

        long res = 1;
        int two = 0;
        for (int v : hm.values()) {
            res *= fact[v];
            two += sum[v];
            res %= m;
        }

        for (int i = 0; i < n; i++) {
            if (a[i] == b[i]) {
                two--;
            }
        }

        res *= powMod(2, two, m);
        res %= m;
        out(res);
    }

    public long powMod(long N, long M, long MOD) {//N^M % MOD
        if(M == 0L)
            return 1L;
        long[] hp = new long[64];
        boolean[] bp = new boolean[64];
        hp[0] = N;
        for(int i = 1; i < hp.length; i++) {
            hp[i] = (hp[i - 1] * hp[i - 1]) % MOD;
        }
        for(int j = 0; j < hp.length; j++) {
            if((M & (1L << j)) != 0)
                bp[j] = true;
        }
        long res = 1;
        for(int i = 0;i < bp.length; i++){
            if(bp[i]) {
                res = (res * hp[i]) % MOD;
            }
        }
        return res;
    }

    void shuffle(int[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            int tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }

    int gcd(int a, int b) {
        while(a != 0 && b != 0) {
            int c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }

    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    public CFA() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new CFA();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}

