Plant

忽然觉得数论的题目模板套的好爽

本来想着用上一题的构造序列的感觉做，莫名翻车，发现没有必要，他只是给你这样好理解的而已

{3   1}*{f（n-1）上}={f（n）上}

 1    3   f（n-1）下     f（n）下

其实就可以直接用构造出的二维数组的n次方解题

再一次在while（scanf）跌倒

F - Plant

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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 Input

Input

1

Output

3

Input

2

Output

10

Hint

The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.

12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061

#include <iostream>#include <cstdio>#include <cstring>#define mod  1000000007using namespace std;struct matrix{    long long ma[3][3];};matrix mult(matrix a,matrix b){    matrix ans;    memset(ans.ma,0,sizeof(ans.ma));    for(int i=1;i<=2;i++)    {        for(int j=1;j<=2;j++)        {            for(int k=1;k<=2;k++)            {                ans.ma[i][j]+=(a.ma[i][k]*b.ma[k][j])%mod;            }            ans.ma[i][j]%=mod;        }    }    return ans;}matrix pow(matrix x,long long n){    matrix ans;    memset(ans.ma,0,sizeof(ans.ma));    for(int i=1;i<=2;i++)    {        ans.ma[i][i]=1;    }    while(n)    {        if(n&1)        {            ans=mult(ans,x);        }        x=mult(x,x);        n>>=1;    }    return ans;}int main(){    long long n;    while(~scanf("%I64d",&n))    {        matrix x;        x.ma[1][1]=x.ma[2][2]=3;        x.ma[2][1]=x.ma[1][2]=1;        x=pow(x,n);        printf("%I64d\n",x.ma[1][1]%mod);    }    return 0;}