1298-one theorem, one year

1298-one theorem, one year

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Time Limit:2 second (s)

Memory limit:32 MB

A number is almost-k-prime if it had exactly K Prime numbers (not necessarily distinct) in its Prime FAC Torization. For example, 2 * 2 * 3 are an almost-3-prime and 2 * 2 * 2 * 2 * 2 are an almost-5-prime number. A number X is called almost-k-first-p-prime If it satisfies the following criterions:

1. X is an almost-k-prime and
2. X have all and only the first P (p≤k) primes in its prime factorization.

For example, if k=3 and p=2, the numbers = 2 * 3 * 3 and * = 2 * 2 * 3 satisfy the above Criteri Ons. and 630 = 2 * 3 * 3 * 5 * 7 is an example of almost-5-first-4-pime.

For a given K and P, your task was to calculate the summation of Φ (X) for all integers x such that's X is an almost-k-first-p-prime.

Input

Input starts with an integer T (≤10000), denoting the number of test cases.

Each case is starts with a line containing, integers K (1≤k≤500) and P (1≤p≤k).

Output

For each case, print the case number and the result modulo 1000000007.

Sample Input	Output for Sample Input
3 3 2 5 4 99 45	Case 1:10 Case 2:816 Case 3:49,939,643

Note

1. In mathematics Φ (X) means the number of relatively prime numbers with respect to X which is smaller tha n X. The numbers relatively prime if their GCD (greatest Common Divisor) is 1. For example, Φ (4), because the numbers, is relatively prime to are:1, 5, 7, 11.
2. For the first case, K = 3 and P = 2 We had only such numbers which is almost-3-first-2-prime , 18=2*3*3 and 12=2*2*3. The result is therefore, Φ (12) +φ (+) = Ten.

Problem Setter:samir ahmedspecial thanks:jane Alam Jan idea: DP; state transition equation Dp[i][j]=dp[i-1][j-1]+dp[i][j-1];i represents the first prime I, J denotes the length of the prime factor consisting of the first prime number of primes, dp[i][j] The existence of all the numbers in accordance with this requirement; State interpretation: The number of the current end is put in the first prime, then its previous number is either its previous prime or himself. So DP first makes a table, and then according to the Euler function N ((1-1/P1) * (1-1/P2) .... Because Dp[i][j] is the same as the number of factors, and then (p1*p2* ...) The table dozen good an, put (p1-1) * (p2-1) *....bn play well so k=i,p=j, the straight is dp[j][i]* (An[j]/bn[j])%mod, and then Bn[i] with Fermat theorem converted to inverse, so finally dp[j][i]* (an[ J]*BN[J])%mod.

1#include <math.h>2#include <stdlib.h>3#include <stdio.h>4#include <algorithm>5#include <iostream>6#include <string.h>7#include <vector>8#include <map>9#include <math.h>Ten using namespacestd; OnetypedefLong LongLL; Atypedef unsignedLong Longll; - BOOLprime[ the]= {0}; - intsu[ -]; theLL dp[ -][ -]; -LL ola[ -]; -LL ola1[ -]; - ConstLL mod=1e9+7; +LL Quick (intNintm); - intMainvoid) + { A         inti,j,k,p,q; at          for(i=2; i<= -; i++) -         { -                  for(J=i; i*j<= the; J + +) -                 { -prime[i*j]=true; -                 } in         } -         intans=1; to          for(i=2; i<= the; i++) +         { -                 if(!Prime[i]) the                 { *su[ans++]=i; $                 }Panax Notoginseng         } -Memset (DP,0,sizeof(DP)); thedp[0][0]=1; +dp[1][1]=2; A          for(i=1; i<= -; i++) the         { +                  for(J=i; j<= -; J + +) -                 { $Dp[i][j]= (((dp[i][j-1]+dp[i-1][j-1]) (%mod) * (Su[i]))%MoD; $                 } -         } -ola[1]=su[1]; theola1[1]=su[1]-1; -          for(i=2; i<= -; i++)Wuyi         { theOla[i]= (su[i]*ola[i-1])%MoD; -Ola1[i]= (su[i]-1) *ola1[i-1]%MoD; Wu         } -          for(i=1; i<= -; i++) About         { $Ola[i]=quick (ola[i],mod-2); -         } -scanf"%d",&k); -         ints; A          for(s=1; s<=k; s++) +         { thescanf"%d%d",&p,&q); -LL cnt=Dp[q][p]; $LL cns=Ola[q]; theLL bns=Ola1[q]; theLL sum= ((CNT*CNS)%mod*bns)%MoD; theprintf"Case %d:", s); theprintf"%lld\n", sum); -         } in         return 0; the } theLL Quick (intNintm) About { theLL ans=1; theLL n=N; the          while(m) +         { -                 if(m&1) the                 {Bayians= (ans*n)%MoD; the                 } then= (n*n)%MoD; -M/=2; -         } the         returnans; the}

1298-one theorem, one year