Let us define two functions *f* and *g* on positive integer numbers. 

<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/120df92771639f1f51a4b216dccd3ed7c32bb859.png" style="max-width: 100.0%;max-height: 100.0%;"/> 

You need to process *Q* queries. In each query, you will be given three integers *l*, *r* and *k*. You need to print the number of integers *x* between *l* and *r* inclusive, such that *g*(*x*) = *k*.
The first line of the input contains an integer *Q* (1 ≤ *Q* ≤ 2 × 10<sup class="upper-index">5</sup>) representing the number of queries. 

*Q* lines follow, each of which contains 3 integers *l*, *r* and *k* (1 ≤ *l* ≤ *r* ≤ 10<sup class="upper-index">6</sup>, 1 ≤ *k* ≤ 9).
In the first example:
 -  *g*(33) = 9 as *g*(33) = *g*(3 × 3) = *g*(9) = 9 -  *g*(47) = *g*(48) = *g*(60) = *g*(61) = 6 -  There are no such integers between 47 and 55. -  *g*(4) = *g*(14) = *g*(22) = *g*(27) = *g*(39) = *g*(40) = *g*(41) = *g*(58) = 4
4
22 73 9
45 64 6
47 55 7
2 62 4
1
4
0
8
4
82 94 6
56 67 4
28 59 9
39 74 4
3
1
1
5
import sys
from itertools import permutations
#a=list(map(int,input().strip().split(' ')))
#n,k,s= map(int, sys.stdin.readline().split(' '))
def prod(i):
    s=str(i)
    pro=1
    for j in s:
        v=(ord(j)-48)
        if(v!=0):pro*=v
    if(pro>9):
        if(pro%10 !=0) :return((pro)%10)
        else:
            return(pro//(pow(10,len(str(pro))-1)))
    else: return pro
    
a=[]
n=1000000
for i in range(1,n+1):
    a.append(prod(i))


q=int(input())
for i in range(q):
    l,r,k= map(int, sys.stdin.readline().split(' '))
    ans=a[l-1:r]
    print(ans)
    print(ans.count(k))
