#include<bits/stdc++.h> using namespace std;

typedef long long ll;

ll mod = 1000000007;

ll quick_pow(ll a, ll b, ll p) {//快速幂求2^n-1

ll ans = 1;
while (b) {
	if (b & 1) ans = (ans * a) % p;
	a = (a * a) % p;
	b >>= 1;
}
return ans;

}//考察了一点二项式定理

int main() {//答案为1cn1+2cn2+3cn3+....ncnn,合并成ans下面的公式

ll n;
cin >> n;
ll ans = n*quick_pow(2, n - 1, mod)%mod;//ans=n*2^n-1
cout << ans;
return 0;

}