Plant（矩阵快速幂）

A. Plant

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, coutstreams or the %I64d specifier.

Output

Print a single integer — the remainder of dividing the number of plants that will point "upwards" in n years by1000000007 (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.

       题意：

       给出 n（0 ~ 10 ^ 18），代表一个正方形经过 n 次的切割后，得到正三角形的个数是多少。

 

       思路：

       矩阵快速幂。递推式子：

       Ui+1 = 3 X Ui + Di；

       Di+1 = Ui + 3 X Di；得出的矩阵就不给出了。

 

       AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

const ll MOD = 1000000007;

mat mul (mat a, mat b) {
    mat c(a.size(), vec(b[0].size()));

    for (int i = 0; i < a.size(); ++i) {
        for (int j = 0; j < b[0].size(); ++j) {
            for (int k = 0; k < b.size(); ++k) {
                c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % MOD;
            }
        }
    }

    return c;
}

mat pow (mat a, ll n) {
    mat b(a.size(), vec(a[0].size()));
    for (int i = 0; i < a.size(); ++i)
        b[i][i] = 1;

    while (n > 0) {
        if (n & 1) b = mul(b, a);
        a = mul(a, a);
        n >>= 1;
    }

    return b;
}

int main() {
    ll n;

    while (~scanf("%I64d", &n)) {
        mat a(2, vec(2));
        a[0][0] = a[1][1] = 3;
        a[0][1] = a[1][0] = 1;
        a = pow(a, n);
        printf("%I64d\n", a[0][0]);
    }

    return 0;
}