#include <bits/stdc++.h>
using namespace std;
void read() {
  freopen("input.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
bool voi(char c) {
  return (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u');
}
long long log_pow(long long b, long long p) {
  if (p == 0) return 1;
  if (p == 1) return b;
  if (p % 2) return b * log_pow(b, p - 1);
  long long r = log_pow(b, p / 2);
  return r * r;
}
int d5l_i() {
  int x;
  scanf("%d", &x);
  return x;
}
long long d5l_l() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char d5l_c() {
  char x;
  scanf("%c", &x);
  return x;
}
long long fact(int n) {
  long long fc = 1;
  for (int i = 2; i <= n; i++) fc *= i;
  return fc;
}
int len(long long n) {
  int c = 0;
  while (n) {
    n /= 2;
    c++;
  }
  return c;
}
bool is_prime(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
const int N = 1e6 + 10;
int cm[10][N], a[N];
vector<int> v;
bool contain(int n) {
  while (n) {
    if (n % 10 == 0) return true;
    n /= 10;
  }
  return false;
}
int myFun(int n) {
  int x = 1;
  while (n) {
    if ((n % 10)) x *= (n % 10);
    n /= 10;
  }
  if (x < 10)
    return x;
  else
    myFun(x);
}
void full() {
  for (int i = 1; i < N; i++) a[i] = myFun(i);
}
void preSum() {
  for (int i = 1; i < 10; i++) {
    for (int j = 1; j < N; j++) cm[i][j] += (cm[i][j - 1] + (a[j] == i));
  }
}
int main() {
  full();
  preSum();
  int n = d5l_i();
  while (n--) {
    int l = d5l_i(), r = d5l_i(), x = d5l_i();
    printf("%d\n", cm[x][r] - cm[x][l - 1]);
  }
}

### Prompt

Create a solution in CPP for the following problem:

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

<image>

<image>

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void read() {
  freopen("input.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
bool voi(char c) {
  return (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u');
}
long long log_pow(long long b, long long p) {
  if (p == 0) return 1;
  if (p == 1) return b;
  if (p % 2) return b * log_pow(b, p - 1);
  long long r = log_pow(b, p / 2);
  return r * r;
}
int d5l_i() {
  int x;
  scanf("%d", &x);
  return x;
}
long long d5l_l() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char d5l_c() {
  char x;
  scanf("%c", &x);
  return x;
}
long long fact(int n) {
  long long fc = 1;
  for (int i = 2; i <= n; i++) fc *= i;
  return fc;
}
int len(long long n) {
  int c = 0;
  while (n) {
    n /= 2;
    c++;
  }
  return c;
}
bool is_prime(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
const int N = 1e6 + 10;
int cm[10][N], a[N];
vector<int> v;
bool contain(int n) {
  while (n) {
    if (n % 10 == 0) return true;
    n /= 10;
  }
  return false;
}
int myFun(int n) {
  int x = 1;
  while (n) {
    if ((n % 10)) x *= (n % 10);
    n /= 10;
  }
  if (x < 10)
    return x;
  else
    myFun(x);
}
void full() {
  for (int i = 1; i < N; i++) a[i] = myFun(i);
}
void preSum() {
  for (int i = 1; i < 10; i++) {
    for (int j = 1; j < N; j++) cm[i][j] += (cm[i][j - 1] + (a[j] == i));
  }
}
int main() {
  full();
  preSum();
  int n = d5l_i();
  while (n--) {
    int l = d5l_i(), r = d5l_i(), x = d5l_i();
    printf("%d\n", cm[x][r] - cm[x][l - 1]);
  }
}

```