我用了快速幂呀！

#include <iostream>
#include <cmath>
using namespace std;
typedef unsigned long long ll;
ll qpow(ll a, ll n)
{
    ll re = 1;
    while(n)
    {
        if(n & 1)
            re = (re * a);
        n >>= 1;
        a = (a * a);
    }
    return re;
}
int main() {
	unsigned long long n;
	cin >> n;
	unsigned long long s=qpow(2,n-1);
	s%=911451407;
	n%=911451407;
	cout << s*n%911451407<< endl;
	return 0;
}