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#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

vector<int> t;
int odp[10][10]{
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0} ,
{2 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0},
{4, 2, 0, 0, 0, 0, 0, 0, 0, 0} ,
{8, 16, 0, 0, 0, 0, 0, 0, 0, 0} ,
{16, 100, 4, 0,0, 0, 0, 0, 0, 0} ,
{32, 616, 72, 0, 0, 0, 0, 0, 0, 0}, 
{64, 4024, 952, 0, 0, 0, 0, 0, 0, 0}, 
{128, 28512, 11680, 0, 0, 0, 0, 0, 0, 0},
{256, 219664, 142800, 160, 0, 0, 0, 0, 0, 0}, 
{512, 1831712, 1788896, 7680, 0, 0, 0, 0, 0, 0}
};
bool czy[20];

int sprawdz(vector<int> v)
{
	int ind = 0;
	while(v.size() > 1)
	{
		ind++;
		if(v[0] < v[1])
			czy[0] = 1;
		if(v[v.size() - 1] < v[v.size() - 2])
			czy[v.size() - 1] = 1;
		for(int i = 1; i < v.size() - 1; i++)
			if(v[i] < v[i - 1] || v[i] < v[i + 1])
				czy[i] = 1;
		for(int i = v.size() - 1; i >= 0; i--)
		{
			if(czy[i])
				v.erase(v.begin() + i);
			czy[i] = 0;
		}
	}
	return ind;
}

int hm(int N)
{
	int licz = 0;
	while(N > 1)
	{
		licz++;
		if(N % 2)
			N = N / 2 + 1;
		else
			N /= 2;
	}
	return licz++;
}

int main()
{
	int N, K;
	LL P;
	cin >> N >> K >> P;
	if(N <= 10 && K <= 10)
		cout << odp[N - 1][K - 1];
	else if(hm(N) < K)
		cout << 0;
	else if(N!= 1 && K == 1)
	{
		LL w = 1;
		for(int i = 1; i < N; i++)
			w = (w * 2) % P;
		cout << w;
	}
	else{
	int w = 0;
	for(int i = 0; i < N; i++)
		t.push_back(i + 1);
	do{
		w+=(sprawdz(t) == K ? 1:0);
	}while(next_permutation(t.begin(), t.begin()+N));
	cout << w;
	}
}