/*
思路:斐波那契数列
*/
#include <iostream>
using namespace std;

int f(int n){
    if(n == 1 || n == 2){
        return 1;
    }

    int m1 = 1;
    int m2 = 1;
    int sum = 0;
    for(int i = 3; i <= n; ++i){
        sum = m1 + m2;
        m1 = m2;
        m2 = sum;
    }
    
    return sum;
}

int main() {
    int n;
    cin >> n;

    cout << f(n) << endl;
}
// 64 位输出请用 printf("%lld")