Lucky is fond of Number theory, one day he was solving a problem related to Euler Totient Function (phi) and found an interesting property of phi : phi(1) = 1, and for x > 1: phi(x) < x.

So if we define a sequence with a₀ = x, and for n > 0: a_n = phi(a_n-1), this sequence will be constant equal to 1 starting from some point. Lets define depth(x) as minimal n such that a_n = 1. 

Now he is wondering how many numbers in a given range have depth equal to given number k. As you are a good programmer help Lucky with his task.

Input

Your input will consist of a single integer T  followed by a newline and T test cases.

Each test cases consists of a single line containing integers m, n, and k.

Output

Output for each test case one line containing the count of all numbers whose depth equals to k in given range [m, n].

Constraints

T < 10001
1 ≤ m ≤ n ≤ 10⁶
0 ≤ k < 20

Example

Input:
5
1 3 1
1 10 2
1 10 3
1 100 3
1 1000000 17

Output:
1
3
5
8
287876

Explanation: suppose number is 5 ; its depth will be 3. ( 5 → 4 → 2 → 1 )

Note: Depth for 1 is 0.

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	sankalp_7: 2021-07-08 14:51:55 I pre-computed in 2D array. But my running time was 1.41 s :( , other people have less than 0.2 s. I think there is a better way.
	subhashis_cse: 2020-07-18 17:32:24 AC in 1 go :-D
	poojitha_792: 2020-06-26 10:30:45 @lakshman can u pls look at my code i am getting tle
	van_persie9: 2018-05-24 23:08:11 @Lakshman can you please tell where I need to optimize my code? I am getting TLE ==(Lakshman)==>Your query part is taking time and in the worst case it is O(n). Last edit: 2018-06-08 08:33:44
	jha4032: 2018-03-22 03:15:31 just precalculate ...... and ............. AC(0.08 sec) Last edit: 2018-03-22 03:16:08
	satyampnc: 2017-10-12 09:34:55 huh..finally AC in 0.18sec after 4 WA.....!!! Last edit: 2017-10-12 09:35:54
	jayharsh: 2017-08-18 09:25:48 Too many times TLE but after get the AC......precomputation is the best
	jayharsh: 2017-08-15 21:48:26 @Lakshman please check my approach.....it is giving TLE ==Lakshman==> In worst for each query it is taking $O(n)$. Here you need $O(log n) $for each query. Last edit: 2018-01-26 14:03:35
	congru_mod: 2017-07-03 12:11:55 finally AC after so many WA!!!!!
	vivace: 2016-12-11 08:41:32 precomputation at its best :)