olion August will eat every thing he has found.

    Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.

fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise

    He gives you 5 numbers n,a,b,c,p,and he will eat fn foods.But there are only p foods,so you should tell him fn

mod p.

Input

The first line has a number,T,means testcase.

    Each testcase has 5 numbers,including n,a,b,c,p in a line.

    1≤T≤10,1≤n≤1018,1≤a,b,c≤109,p is a prime number,and p≤109+7

.

Output

Output one number for each case,which is fn

mod p.

Sample Input

1
5 3 3 3 233

Sample Output

190

題解：看到表示式 很容易我們可以想到 取一下 log  然後就可以得到 f[n] = c*f[n - 1] + f[n - 2] + b;  然後構造矩陣就可以了

比較坑的是，a%p==0 的時候要判斷一下，因為 n>=2  的每一項取模之後都為0了 但是 在快速冪中 是無法得到0 的

#include<iostream>
#include<cstdio>
typedef long long ll;
using namespace std;
ll n,a,b,c,p;
struct node{
	ll mat[4][4];
}ans,res;
node cul(node x,node y)
{
	node z;
	for(int i=1;i<=3;i++)
	{
		for(int j=1;j<=3;j++)
		{
			z.mat[i][j]=0;
			for(int k=1;k<=3;k++)
				z.mat[i][j]=(z.mat[i][j]+x.mat[i][k]*y.mat[k][j]%(p-1))%(p-1);
		}
	}
	return z;
}
ll J_ksm()
{
	ll m=n-2;
	while(m)
	{
		if(m&1) res=cul(res,ans);
		m>>=1;
		ans=cul(ans,ans);
	}
//	cout<<res.mat[1][1]<<" "<<res.mat[1][3]<<endl;;
	return (res.mat[1][1]*b%(p-1)+res.mat[1][3]*b%(p-1))%(p-1);
}
ll ksm(ll x,ll y)
{
	ll sum=1;
	while(y)
	{
		if(y&1) sum=(sum*x)%p;
		y>>=1;
		x=x*x%p;
	}
	return sum;
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);
		ans.mat[1][1]=c,ans.mat[1][2]=1,ans.mat[1][3]=1;
		ans.mat[2][1]=1,ans.mat[2][2]=0,ans.mat[2][3]=0;
		ans.mat[3][1]=0,ans.mat[3][2]=0,ans.mat[3][3]=1;
		for(int i=1;i<=3;i++)
		{
			for(int j=1;j<=3;j++)
			{
				if(i==j) res.mat[i][j]=1;
				else res.mat[i][j]=0;
			}
		}
		if(n==1) printf("1\n");
		else if(n==2) printf("%lld\n",ksm(a,b));
		else if(a%p==0) printf("0\n");
		else
		{
			ll cnt=J_ksm();
		//	cout<<cnt<<endl;
			printf("%lld\n",ksm(a,cnt));
		}
	}
	return 0;
}