Plant （矩阵快速幂）

题目链接：http://codeforces.com/problemset/problem/185/A

题目：

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).

Examples

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：就是初始时是一个朝向的三角形，每年都会把一个三角形分成四个，三个与原来的那个三角形的朝向（上下）相同，一个不同，问第n年时有多少个三角形朝上~

思路：对于初始的矩阵f[0]为朝上的三角形个数，f[1]为朝下的三角形个数。通过对n=1,2,3进行列举可以推出转移矩阵为a[0][0] = 3, a[0][1] = 1， a[1][0] = 1, a[1][1] = 3。

代码实现如下：

 1 #include <cstring>
 2 #include <iostream>
 3 using namespace std;
 4 
 5 typedef long long ll;
 6 const int mod = 1e9 + 7;
 7 
 8 ll n;
 9 int f[2], a[2][2];
10 
11 void mul(int f[2], int a[2][2]) {
12     int c[2];
13     memset(c, 0, sizeof(c));
14     for(int i = 0; i < 2; i++) {
15         for(int j = 0; j < 2; j++) {
16             c[i] = (c[i] + (ll) f[j] * a[j][i]) % mod;
17         }
18     }
19     memcpy(f, c, sizeof(c));
20 }
21 
22 void mulself(int a[2][2]) {
23     int c[2][2];
24     memset(c, 0, sizeof(c));
25     for(int i = 0; i < 2; i++) {
26         for(int j = 0; j < 2; j++) {
27             for(int k = 0; k < 2; k++) {
28                 c[i][j] = (c[i][j] + (ll) a[i][k] * a[k][j]) % mod;
29             }
30         }
31     }
32     memcpy(a, c, sizeof(c));
33 }
34 
35 int main() {
36     while(cin >>n) {
37         f[0] = 1, f[1] = 0;
38         a[0][0] = 3, a[0][1] = 1;
39         a[1][0] = 1, a[1][1] = 3;
40         for(; n; n >>= 1) {
41             if(n & 1) mul(f, a);
42             mulself(a);
43         }
44         cout << (f[0] % mod) <<endl;
45     }
46     return 0;
47 }