Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition:

• T is a subset of S
• |T| = m
• T does not contain continuous numbers, that is to say x and x+1 can not both in T

Input

There are multiple cases, each contains 3 integers n ( 1 <= n

Output

Output the total number mod p.

Sample Input

5 1 11
5 2 11

Sample Output

5
6

題意： 一共有n個元素，選擇m個不相鄰的元素，一共有多少種選法 分析： 用插板法求出組合數。從n個元素中選出m個，剩下n-m個元素產生n-m+1個空位，把m個元素插到n-m+1個空位，方法數為C(n-m+1,m);

#include<cstdio>
using namespace std;
typedef long long ll;
ll fastPow(ll a,ll b,ll p)
{
    ll ret=1;
    while(b){
        if(b&1) ret=(ret*a)%p;
        a=(a*a)%p;
        b>>=1;
    }
    return ret;
}
ll c(ll n,ll m,ll p)
{

    if(m==0)    return 1;
    //if(m>n-m)   m=n-m;
    ll up=1,down=1;
    for(int i=1;i<=m;i++)
    {
        up=up*(n-i+1)%p;
        down=(down*i)%p;
    }
    return up*fastPow(down,p-2,p)%p;
}

ll lucas(ll n,ll m,ll p)
{
    if(m==0)    return 1;
    return c(n%p,m%p,p)*lucas(n/p,m/p,p)%p;
}
/*
ll lucas(ll n,ll m,ll p)
{
    ll ans=1;
    while(n&&m&&ans){
        ans=ans*c(n%p,m%p,p)%p;
        n/=p;
        m/=p;
    }
    return ans;
}
*/
int main()
{
    ll n,m,p;
    while(scanf("%lld%lld%lld",&n,&m,&p)!=EOF){
        n=n-m+1;
        if(n<m){
            printf("0\n");
            continue;
        }
        if(n==m)
        {
        printf("1\n");
        continue;
        }
        if(m>n-m)
            m=n-m;
        printf("%lld\n",lucas(n,m,p));
    }
    return 0;
}