C

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231231 as described above.

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

Sample Input

2
3 1 2
4 1 10

Sample Output

85
369

Hint
In the first case, the third number is 85 = 2*1十2十3^4.
In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

 a[i] = a[i-1] + 2*a[i-2] + i^4    利用矩阵快速幂,,维护一个a[i]  a[i-1] i^4 i^3 i^2 i 1 构造出矩阵

难点在于如何构造i^4 利用递推构造 i^3 i^2 i 1,就可以得到答案

#include<bits/stdc++.h>using namespace std;typedef long long ll;const ll MOD = 2147493647;struct matrix{    ll x[8][8];};matrix mutimatrix(matrix a,matrix b){    matrix temp;    memset(temp.x,0,sizeof(temp.x));    for(int i=0;i<7;i++)    {        for(int j=0;j<7;j++)        {            for(int k=0;k<7;k++)            {                temp.x[i][j] += ( (a.x[i][k]%MOD)*(b.x[k][j]%MOD) )%MOD;                temp.x[i][j] %= MOD;            }        }    }    return temp;}matrix powmatr(matrix a,ll n){    matrix temp;    memset(temp.x,0,sizeof(temp.x));    for(int i=0;i<7;i++)    {        temp.x[i][i]=1;    }    while(n)    {        if(n&1)        {            temp = mutimatrix(temp,a);        }        a = mutimatrix(a,a);        n>>=1;    }    return temp;}int main(){//    freopen("data.txt","r",stdin);//    ios_base::sync_with_stdio(false);    int T;    ll a,b,n;    matrix cur;    scanf("%d",&T);    while(T--)    {        memset(cur.x,0,sizeof(cur.x));        cur.x[0][0]=1;        cur.x[0][1]=2;        cur.x[0][2]=1;        cur.x[0][3]=4;        cur.x[0][4]=6;        cur.x[0][5]=4;        cur.x[0][6]=1;        cur.x[1][0]=1;        cur.x[1][1]=0;        cur.x[1][2]=0;        cur.x[1][3]=0;        cur.x[1][4]=0;        cur.x[1][5]=0;        cur.x[1][6]=0;        cur.x[2][0]=0;        cur.x[2][1]=0;        cur.x[2][2]=1;        cur.x[2][3]=4;        cur.x[2][4]=6;        cur.x[2][5]=4;        cur.x[2][6]=1;        cur.x[3][0]=0;        cur.x[3][1]=0;        cur.x[3][2]=0;        cur.x[3][3]=1;        cur.x[3][4]=3;        cur.x[3][5]=3;        cur.x[3][6]=1;        cur.x[4][0]=0;        cur.x[4][1]=0;        cur.x[4][2]=0;        cur.x[4][3]=0;        cur.x[4][4]=1;        cur.x[4][5]=2;        cur.x[4][6]=1;        cur.x[5][0]=0;        cur.x[5][1]=0;        cur.x[5][2]=0;        cur.x[5][3]=0;        cur.x[5][4]=0;        cur.x[5][5]=1;        cur.x[5][6]=1;        cur.x[6][0]=0;        cur.x[6][1]=0;        cur.x[6][2]=0;        cur.x[6][3]=0;        cur.x[6][4]=0;        cur.x[6][5]=0;        cur.x[6][6]=1;        scanf("%lld%lld%lld",&n,&a,&b);        if(n==1)        {            printf("%lld\n",a%MOD);            continue;        }        if(n==2)        {            printf("%lld\n",b%MOD);            continue;        }        cur = powmatr(cur,n-2);        matrix ans;        memset(ans.x,0,sizeof(ans.x));        ans.x[0][0]=b%MOD;        ans.x[1][0]=a%MOD;        ans.x[2][0]=16;        ans.x[3][0]=8;        ans.x[4][0]=4;        ans.x[5][0]=2;        ans.x[6][0]=1;        ans = mutimatrix(cur,ans);        printf("%lld\n",(ans.x[0][0]+MOD)%MOD);    }    return 0;}