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

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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. 

Input

The first line of the input contains an integer Q (1 ≤ Q ≤ 2 × 105) representing the number of queries. 

Q lines follow, each of which contains 3 integers l, r and k (1 ≤ l ≤ r ≤ 106, 1 ≤ k ≤ 9).

Output

For each query, print a single line containing the answer for that query.

Examples

Input

4
22 73 9
45 64 6
47 55 7
2 62 4


Output

1
4
0
8


Input

4
82 94 6
56 67 4
28 59 9
39 74 4


Output

3
1
1
5

Note

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long a[1000010], b[11][1000010 + 1];
long long power(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long i, j;
  for (i = 1; i <= 9; i++) {
    a[i] = i;
  }
  for (i = 10; i < 1000010; i++) {
    long long x = i;
    long long c = 1;
    while (x) {
      long long y = x % 10;
      if (y) c *= y;
      x /= 10;
    }
    a[i] = a[c];
  }
  for (i = 0; i < 10; i++) {
    b[i][0] = 0;
  }
  for (j = 1; j < 10; j++) {
    long long c = 0;
    for (i = 1; i < 1000010; i++) {
      if (a[i] == j) c++;
      b[j][i] = c;
    }
  }
  int q;
  cin >> q;
  while (q--) {
    long long l, r, k;
    cin >> l >> r >> k;
    long long ans = b[k][r] - b[k][l - 1];
    printf("%lld\n", ans);
  }
  return 0;
}