Plant

Dwarfs have planted a very interesting plant, which is a triangle directed “upwards”. This plant has an amusing feature. After one year a triangle plant directed “upwards” divides into four triangle plants: three of them will point “upwards” and one will point “downwards”. After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.

Help the dwarfs find out how many triangle plants that point “upwards” will be in n years.

Input
The first line contains a single integer n (0 ≤ n ≤ 1018) — the number of full years when the plant grew.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output
Print a single integer — the remainder of dividing the number of plants that will point “upwards” in n years by 1000000007 (109 + 7).

Sample test(s)
input
1
output
3
input
2
output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
每次按图示生出小三角形，问最后出来多少向上的三角形。
设an是n次后向上的三角形，bn是向下的个数，那么可以发现，一个向上可以生出3个向上，一个向下可以生出1个向上，就是
an = 3*a(n-1) + b(n-1);
又三角形个数是一定的，就是an + bn 是4的n-1次方，就可以化简第一个式子，下面就是推导表达式了，高中数学题，过程电脑不好写，反正我会写。
之后就是一个快速幂。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const ll mod = 1000000007;
ll pow_mod(ll a, ll n)
{
    if (n== 0)return 1;
    ll x = pow_mod(a, n / 2);
    ll ans = x*x%mod;
    if (n % 2 == 1)
        ans = ans*a%mod;
    return ans;
}
int main()
{
    ll i, j, m, n, ans, t;
    cin >> n;
    if (n == 0)
        cout << "1" << endl;
    else
        cout << (pow_mod(2, 2 * n - 1) + pow_mod(2, n - 1)) % mod << endl;
    return 0;
}