#include <stdio.h>
#include <algorithm>
#include <vector>
#include <deque>
#include <string>
#include <iostream>
using namespace std;
using ll = long long;


template<class T = long long, T init = 0>
class Mat {
private:
	T mul(T a, T b) const {
		return a * b % mod;
	}
	T sum(T a, T b) const {
		return (a + b) % mod;
	}
public:
	int n, m, mod;
	vector<vector<ll>> data;
	Mat(int n, int m, int mod) :n(n), m(m), mod(mod), data(n, vector<T>(m, init)) {};

	Mat operator+(Mat& o) const {
		Mat r = Mat(n, m, mod);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				r.data[i][j] = sum(data[i][j], o.data[i][j]);

	}
	Mat operator*(const Mat& o) const {
		Mat r = Mat(n, o.m, mod);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < o.m; j++)
				for (int k = 0; k < m; k++)
					r.data[i][j] = sum(r.data[i][j], mul(data[i][k], o.data[k][j]));
		return r;
	}
};

int n, m;
Mat<> bpow(Mat<> b, int e) {
	Mat<> r = Mat<>(2, 2, m);
	r.data[0][0] = r.data[1][1] = 1;
	while (0 < e) {
		if (e & 1)
			r = r * b;
		b = b * b;
		e /= 2;
	}
	return r;
}
int main() {
	scanf("%d %d", &n, &m);
	Mat<> k(2, 2, m);
	k.data[0][0] = k.data[0][1] = k.data[1][0] = 1;
	Mat<> u(2, 1, m);
	u.data[0][0] = 1;
	ll r = (bpow(k, n - 2) * u).data[0][0];
	printf("%lld\n", r);
}