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A big marathon is held on Al-Maza Road, Damascus. Runners came from all over the world to run all the way along the road in this big marathon day. The winner is the player who crosses the finish line first.
The organizers write down finish line crossing time for each player. After the end of the marathon, they had a doubt between 2 possible winners named "Player1" and "Player2". They will give you the crossing time for those players and they want you to say who is the winner?
First line contains number of test cases (1 ≤ T ≤ 100). Each of the next T lines represents one test case with 6 integers H1 M1 S1 H2 M2 S2. Where H1, M1, S1 represent Player1 crossing time (hours, minutes, seconds) and H2, M2, S2 represent Player2 crossing time (hours, minutes, seconds). You can assume that Player1 and Player2 crossing times are on the same day and they are represented in 24 hours format(0 ≤ H1,H2 ≤ 23 and 0 ≤ M1,M2,S1,S2 ≤ 59)
H1, M1, S1, H2, M2 and S2 will be represented by exactly 2 digits (leading zeros if necessary).
For each test case, you should print one line containing "Player1" if Player1 is the winner, "Player2" if Player2 is the winner or "Tie" if there is a tie.
3 18 03 04 14 03 05 09 45 33 12 03 01 06 36 03 06 36 03
Player2 Player1 Tie
1 #include2 using namespace std; 3 typedef long long ll; 4 inline int read() 5 { 6 int x=0,f=1; 7 char ch=getchar(); 8 while(ch<'0'||ch>'9') 9 {10 if(ch=='-')11 f=-1;12 ch=getchar();13 }14 while(ch>='0'&&ch<='9')15 {16 x=x*10+ch-'0';17 ch=getchar();18 }19 return x*f;20 }21 inline void write(int x)22 {23 if(x<0)24 {25 putchar('-');26 x=-x;27 }28 if(x>9)29 {30 write(x/10);31 }32 putchar(x%10+'0');33 }34 int a[3],b[3];35 int n;36 int main()37 {38 n=read();39 while(n--)40 {41 //cin.getline(s1,8);42 //cin.getline(s2,8);43 for(int i=0;i<3;i++)44 a[i]=read();45 for(int i=0;i<3;i++)46 b[i]=read();47 int flag=-1;48 for(int i=0;i<3;i++)49 {50 if(a[i]>b[i])51 {52 flag=1;53 break;54 }55 else if(a[i]
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