-----Problem Statement-----
Levy's conjecture, named after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessary to be distinct) for n > 2. (Source: Wikipedia)
In this problem, given a positive integer N (not necessary to be odd integer greater than 5). Your task is to calculate how many distinct ordered pairs (p, q) such that N = p + 2q, where p and q are primes.
-----Input-----
The first line of input contains an integer T, denoting the number of test cases. Then T test cases follow.
Each test case consists of exactly one line containing an integer N.
-----Constraints-----
- 1 ≤ T ≤ 100000 (105)
- 1 ≤ N ≤ 10000 (104)
-----Output-----
For each test case, output the number of ordered pairs (p, q) of primes such that N = p + 2q.
-----Example-----
Input:
3
2
7
11
Output:
0
1
2
-----Explanation-----
Case #1: There are no ordered pairs (p, q) such that p + 2q = 2.
Case #2: There is only one ordered pair (p, q) = (3, 2) such that p + 2q = 7.
Case #3: There are two ordered pairs (p, q) = (7, 2), (5, 3) such that p + 2q = 11.
import math
MAX = 10001
PRIME=[]
A=[1 for i in range(MAX)]
A[0] = 0
A[1] = 0
B=[0 for i in range(MAX)]

for i in range(2,int(math.sqrt(MAX))+1):
 if A[i] is 1:
  for j in range(i*i,MAX,i):
   A[j] = 0
for i in range(MAX):
 if A[i] is 1:
  PRIME.append(i)

for p in PRIME:
 for q in PRIME:
  if p+(2*q)>=MAX:
   break
  B[p+(2*q)]+=1

for _ in range(int(input())):
 n=int(input())
 print(B[n])

  
  

import math
MAX = 10001
PRIME=[]
A=[1 for i in range(MAX)]
A[0] = 0
A[1] = 0
B=[0 for i in range(MAX)]

for i in range(2,int(math.sqrt(MAX))+1):
 if A[i] is 1:
  for j in range(i*i,MAX,i):
   A[j] = 0
for i in range(MAX):
 if A[i] is 1:
  PRIME.append(i)

for p in PRIME:
 for q in PRIME:
  if p+(2*q)>=MAX:
   break
  B[p+(2*q)]+=1

for _ in range(int(input())):
 n=int(input())
 print(B[n])

  
  
