We say that a odd number N is similar to 2017 when both N and (N+1)/2 are prime.

You are given Q queries.

In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Constraints

* 1≤Q≤10^5
* 1≤l_i≤r_i≤10^5
* l_i and r_i are odd.
* All input values are integers.

Input

Input is given from Standard Input in the following format:


Q
l_1 r_1
:
l_Q r_Q


Output

Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th query.

Examples

Input

1
3 7


Output

2


Input

4
13 13
7 11
7 11
2017 2017


Output

1
0
0
1


Input

6
1 53
13 91
37 55
19 51
73 91
13 49


Output

4
4
1
1
1
2
import math
Q = int(raw_input())
L, R = [], []
for i in range(Q):
	l_r = [int(num) for num in raw_input().split()]
	L.append(l_r[0])
	R.append(l_r[1])

max_num = max(R)
def calc_sosus(max_num):
	IsPrimes = [True] * (max_num + 1)
	IsPrimes[0] = False
	IsPrimes[1] = False
	for min_i in range(2, int(math.sqrt(max_num)) + 1):
		if IsPrimes[min_i]:
			for num in range(min_i*min_i, max_num+1, min_i):
				IsPrimes[num] = False
	return IsPrimes

IsPrimes = calc_sosus(max_num)

c_s = [0] * (max_num + 1)
for i in range(2, max_num + 1):
	if (IsPrimes[i]) & (((i+1) % 2)==0) & (IsPrimes[(i+1)/2]): 
		c_s[i] = c_s[i-1] + 1
	else:
		c_s[i] = c_s[i-1]

for i in range(Q):
	print c_s[R[i]] - c_s[L[i]-1]