CodeForces-185A（矩阵快速幂）

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 nyears.

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).

Example

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：你的第一年有一个向上的三角形，然后每一年每个三角形就会产生3个方向与其相同的三角形和一个方向不同的三角形。你的任务是求第n年的向上的三角形个数。数据可能很大，请将个数%10e9+7题解：构造矩阵为| 1  3 | | 3  1 |*/#include <stdio.h>#include <string.h>#include<iostream>#include <algorithm>using namespace std;#define LL __int64struct node{LL maxtri[3][3];};node res, roi;LL n, kk = 1000000007;node operator * (node &a, node &b){node ans;for (int i = 0; i < 2; i++){for (int j = 0; j < 2; j++){ans.maxtri[i][j] = 0;for (int k = 0; k < 2; k++){ans.maxtri[i][j] += a.maxtri[i][k] * (b.maxtri[k][j]);ans.maxtri[i][j] %= kk;}}}return ans;}void quickmuti(LL nn){node temp;for (int i = 0; i < 2; i++){for (int j = 0; j < 2; j++){temp.maxtri[i][j] = (i == j);}}while (nn){if (nn & 1){temp = temp*roi;}nn >>= 1;roi = roi*roi;}res = res*temp;}void init(){memset(roi.maxtri, 0, sizeof(roi.maxtri));memset(res.maxtri, 0, sizeof(res.maxtri));roi.maxtri[0][0] = roi.maxtri[1][1] = 3;roi.maxtri[0][1] = roi.maxtri[1][0] = 1;res.maxtri[0][0] = 1;}int main(){while (scanf("%I64d", &n) != EOF){init();quickmuti(n);printf("%I64d\n", res.maxtri[0][0] % kk);}return 0;}