Problem Description


I've sent Fang Fang around 201314 text messages in almost 5 years. Why can't she make sense of what I mean?
``But Jesus is here!" the priest intoned. ``Show me your messages."
Fine, the first message is  s1=‘‘c" and the second one is  s2=‘‘ff".
The  i-th message is  si=si−2+si−1 afterwards. Let me give you some examples.
s3=‘‘cff",  s4=‘‘ffcff" and  s5=‘‘cffffcff".

``I found the  i-th message's utterly charming," Jesus said.
``Look at the fifth message".  s5=‘‘cffffcff" and two  ‘‘cff" appear in it.
The distance between the first  ‘‘cff" and the second one we said, is  5.
``You are right, my friend," Jesus said. ``Love is patient, love is kind.
It does not envy, it does not boast, it is not proud. It does not dishonor others, it is not self-seeking, it is not easily angered, it keeps no record of wrongs.
Love does not delight in evil but rejoices with the truth.
It always protects, always trusts, always hopes, always perseveres."

Listen - look at him in the eye. I will find you, and count the sum of distance between each two different  ‘‘cff" as substrings of the message.


 



Input


T (1≤T≤100), indicating there are  T test cases.
Following  T lines, each line contain an integer  n (3≤n≤201314), as the identifier of message.


 



Output


T lines.
Each line contains an integer equaling to:

∑i<j:sn[i..i+2]=sn[j..j+2]=‘‘cff"(j−i) mod 530600414,



where 

sn as a string corresponding to the  n-th message.


 



Sample Input


9 5 6 7 8 113 1205 199312 199401 201314


 



Sample Output


Case #1: 5 Case #2: 16 Case #3: 88 Case #4: 352 Case #5: 318505405 Case #6: 391786781 Case #7: 133875314 Case #8: 83347132 Case #9: 16520782


 



递推一下

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
const int maxn=300005;
typedef long long LL;
const LL base=530600414;
int T,t=0,n;

struct point
{
    LL l,r,x,c,len;
    point(){};
    point(LL l,LL r,LL x,LL c,LL len):l(l),r(r),x(x),c(c),len(len){}; 
}f[maxn];

void init()
{
    f[1]=point(0,0,1,0,1);
    f[2]=point(0,0,0,0,2);
    f[3]=point(0,2,1,0,3);
    f[4]=point(2,2,1,0,5);
    for (int i=5;i<=201314;i++)
    {
        f[i].len=(f[i-1].len+f[i-2].len)%base;
        f[i].x=(f[i-1].x+f[i-2].x)%base;
        f[i].l=(f[i-2].l+f[i-1].l+f[i-1].x*f[i-2].len)%base;
        f[i].r=(f[i-1].r+f[i-2].r+f[i-2].x*f[i-1].len)%base;
        f[i].c=(f[i-1].c+f[i-2].c+f[i-1].x*f[i-2].r+f[i-2].x*f[i-1].l+f[i-1].x*f[i-2].x)%base;
    }
}

int main()
{
    init();
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        printf("Case #%d: %d\n",++t,f[n].c);
    }
    return 0;
}