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
#include <bitset>
#include <climits>
#include <cstring>
#include <iostream>
#include <vector>
#include <cmath>
#define ll long long

using namespace std;


int main(){
      int o = pow(10, 6) + 1;
      ll dp[o];

      for(int i = 0; i < 10; i++){
            dp[i] = i;
      }
      
      for(int i = 0; i < o; i++){
            int index = 1;
            ll val = i;
            while(val){
                  if(val%10)
                        index *= val %10;
                  val /= 10;
            }
            dp[i] = dp[index];
      }

      vector<ll> running_count(10,0);

      vector<vector<ll>> count_arr;
      for(int i = 0; i < o; i++){
          running_count[dp[i]] += 1;
          vector<ll> current(running_count);
          count_arr.push_back(current);
      }

      int q; 
      cin >> q;
      for(int j = 0; j < q; j++){

            int l,r,k;
            cin >> l >> r >> k;
            cout << count_arr[r][k] - count_arr[l-1][k] << endl;
      }


}
