An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Problem Description

One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.

1. multiply X with a number.

2. divide X with a number which was multiplied before.

After each operation, please output the number X modulo M.

Input

The first line is an integer T(

Output

For each test case, the first line, please output "Case #x:" and x is the id of the test cases starting from 1.

Then Q lines follow, each line please output an answer showed by the calculator.

Sample Input

1 10 1000000000 1 2 2 1 1 2 1 10 2 3 2 4 1 6 1 7 1 12 2 7

Sample Output

Case #1: 2 1 2 20 10 1 6 42 504 84

Source

题解：考虑到取摸，离线做

或者线段树上搞

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                  using namespace std ;typedef long long ll;#define mem(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,127,sizeof(a));#define TS printf("111111\n");#define FOR(i,a,b) for( int i=a;i<=b;i++)#define FORJ(i,a,b) for(int i=a;i>=b;i--)#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)#define mod 1000000007#define inf 100000inline ll read(){ ll x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-')f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); } return x*f;}//****************************************struct ss{ int id,x; bool operator < (const ss &A)const { return id < A.id; }};#define maxn 100000+5set
                  
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