F - Plant （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 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).

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：问你第n个大三角形中有几个小三角形是朝上的。

思路：每个朝上的小三角形下一年会变成三个朝上一个朝下，每个向下的小三角下年会变成三个朝下一个朝上。
列出几个你会发现朝上的分别为1, 1+2, 1+2+3+4, 1+2…+8
即sum=(（1+2^n）*2^n)/2，就是求等差数列的和的公式：头加尾乘以总个数再除以2。最后就是2^(n-1)+2^(2n-1)

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
LL quick_mul(LL a,LL b){
    a%=mod;
    LL ans=1;
    while(b){
        if(b&1)
            ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
   LL n,sum=0;
   scanf("%I64d",&n);
   if(n==0)
        printf("1\n");
   else{
   sum=(quick_mul(2,n-1)+quick_mul(2,2*n-1))%mod;
   printf("%I64d\n",sum);
   }
   return 0;
}