cf 167.d（ 多重集全排列 ）

高中数学不好的娃实在是伤不起.

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  一个公式：

设多重集 S={n_1×e_{1 ，} n_2×e_{2，... ，}n_k×e_k},令a_n为S的全排列数，则

a_n=(n₁+n₂+...+n_k)! / (n₁!n₂!...n_k!) .

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然后这题就好做了.

D. Dima and Two Sequences

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Dima has two sequences of points with integer coordinates: sequence (a₁, 1), (a₂, 2), ..., (a_n, n) and sequence (b₁, 1), (b₂, 2), ..., (b_n, 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), ..., (p_2·n, q_2·n) and (x₁, y₁), (x₂, y₂), ..., (x_2·n, y_2·n) distinct, if there is such i (1 ≤ i ≤ 2·n), that (p_i, q_i) ≠ (x_i, y_i).

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, ..., a_n (1 ≤ a_i ≤ 10⁹). The third line containsn integers b₁, b₂, ..., b_n (1 ≤ b_i ≤ 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.

Sample test(s)

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.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;

typedef __int64 ll;

struct node
{
    int x,y;
}g[1001000];
ll sum;
int n;
ll MOD;
int cmp(node x,node y)
{
    if(x.x!=y.x) return x.x<y.x;
    else return x.y<y.y;

}

void fuc(int s,int t)
{
    if(t-s==1) return ;
    
    ll cnt=0;
    for(int i=s;i<t;i++)
    {
        if(i==t-1)
        {
            cnt++;
            break;
        }
        if(g[i].y==g[i+1].y)
        {
            i++;
        }
        cnt++;
    }
    ll tmp=t-s-cnt;
    ll ttmp=1;
    for(int i=1;i<=t-s;i++)
    {
        ll ti=i;
        if(i%2==0&&tmp!=0)
        {
            ti=i/2;
            tmp--;
        }
        ttmp=(ttmp*ti)%MOD;
    }
    sum=(sum*ttmp)%MOD;
}
int main()
{
    sum=1;
    int cnt=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        g[cnt].x=tmp;
        g[cnt++].y=i;
    }
    for(int i=1;i<=n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        g[cnt].x=tmp;
        g[cnt++].y=i;
    }
    scanf("%I64d",&MOD);
    sort(g,g+cnt,cmp);
    int tmp=g[0].x;
    int f=0;
    for(int i=1;i<=cnt;i++)
    {
        if(g[i].x!=tmp||i==cnt)
        {
            fuc(f,i);
            if(i==cnt) break;
            f=i;
            tmp=g[i].x;
        }
    }
    printf("%I64d",sum);
    return 0;
}