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;
const double pi = 3.14159265359;
long long max(long long a, long long b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a <= b)
    return a;
  else
    return b;
}
bool valid(int i, int j, long long n, long long m) {
  if (i >= 0 && i < n && j >= 0 && j < m)
    return 1;
  else
    return 0;
}
void yes() { cout << "YES" << '\n'; }
void no() { cout << "NO" << '\n'; }
vector<unsigned long long> divisor(10000007);
void printDivisors(long long n) {
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        divisor[n] += i;
      } else {
        divisor[n] += i;
        divisor[n] += n / i;
      }
    }
  }
}
bool prime[1200000];
map<long long, long long> primes;
void SieveOfEratosthenes(long long n = 1200000) {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (int p = 2; p * p < n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i < n; i += p) prime[i] = false;
    }
  }
  for (int i = 1; i < n + 1; i++)
    if (prime[i]) {
      primes[i]++;
    }
}
int gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
string toBinary(long long n) {
  string r;
  while (n != 0) {
    r = (n % 2 == 0 ? "0" : "1") + r;
    n /= 2;
  }
  return r;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
map<long long, long long> PF;
void primeFactors(long long n) {
  while (n % 2 == 0) {
    PF[2]++;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      PF[i]++;
      n = n / i;
    }
  }
  if (n > 2) PF[n]++;
}
long double intlog(long long base, long long x) {
  long double ans1 = log10(x), ans2 = log10(base);
  return (long double)(ans1 / ans2);
}
bool IsPowerOfTwo(long long x) { return (x & (x - 1)) == 0; }
map<long long, long long> LN;
void GetLuckyNumbers(string s) {
  if (s.size() > 10) return;
  GetLuckyNumbers(s + '4');
  if (s != "") LN[stoll(s)]++;
  GetLuckyNumbers(s + '7');
  if (s != "") LN[stoll(s)]++;
}
unsigned long long toDecimal(unsigned long long n) {
  unsigned long long num = n;
  unsigned long long dec_value = 0;
  unsigned long long base = 1;
  int temp = num;
  while (temp) {
    unsigned long long last_digit = temp % 10;
    temp = temp / 10;
    dec_value += last_digit * base;
    base = base * 2;
  }
  return dec_value;
}
long double power(long double b, long long p) {
  if (p == 0)
    return 1;
  else if (p == 1)
    return b;
  long double a = power(b, p / 2);
  if (p % 2 == 0) return (a) * (a);
  return (b) * (a) * (a);
}
long largest_power_of_two(long N) {
  N = N | (N >> 1);
  N = N | (N >> 2);
  N = N | (N >> 4);
  N = N | (N >> 8);
  return (N + 1) >> 1;
}
long long cnt(string s) {
  long long c = 0;
  for (int i = 0; i < s.size(); i++) c += s[i] - '0';
  return c;
}
bool ispal(string s) {
  for (int i = 0; i < s.size() / 2; i++) {
    if (s[i] != s[s.size() - i - 1]) return false;
  }
  return true;
}
map<string, long long> substrings;
void subString(string s, long long n) {
  for (int i = 0; i < n; i++)
    for (int len = 1; len <= n - i; len++) {
      substrings[s.substr(i, len)]++;
    }
}
bool check_divisabilty_prefix(string str, long long a) {
  long long len = str.size();
  vector<int> lr(len + 1, 0);
  lr[0] = (str[0] - '0') % a;
  for (int i = 1; i < len; i++)
    lr[i] = ((lr[i - 1] * 10) % a + (str[i] - '0')) % a;
  if (lr[len - 1] == 0) return 1;
  return 0;
}
bool check_divisabilty_suffix(string str, long long b) {
  long long len = str.size();
  vector<long long> rl(len + 1, 0);
  rl[len - 1] = (str[len - 1] - '0') % b;
  int power10 = 10;
  for (int i = len - 2; i >= 0; i--) {
    rl[i] = (rl[i + 1] + (str[i] - '0') * power10) % b;
    power10 = (power10 * 10) % b;
  }
  if (rl[0] == 0) return 1;
  return 0;
}
bool isSubSeq(int idx1, int idx2, string s1, string s2) {
  if (idx2 == s2.size()) return true;
  if (idx1 == s1.size()) return false;
  if (s1[idx1] == s2[idx2])
    return isSubSeq(idx1 + 1, idx2 + 1, s1, s2);
  else
    return isSubSeq(idx1 + 1, idx2, s1, s2);
}
vector<long long> v;
long long bs(long long st, long long en, long long rq, long long val) {
  long long ans;
  while (st <= en) {
    long long mid = (st + en) / 2;
    long long aa = v[mid] - v[rq];
    if (aa <= val) {
      st = mid + 1;
      ans = mid;
    } else
      en = mid - 1;
  }
  return abs(ans - rq);
}
bool isPerfectSquare(long double x) {
  if (x >= 0) {
    long long sr = sqrt(x);
    return (sr * sr == x);
  }
  return false;
}
long long fn(long long s) {
  if (s < 10) return s;
  long long pr = 1;
  while (s) {
    if (s % 10) pr *= (s % 10);
    s /= 10;
  }
  return fn(pr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  vector<vector<long long> > v(11, vector<long long>(1e6 + 1, 0));
  for (int i = 1; i <= 1000000; i++) {
    v[fn(i)].push_back(i);
  }
  int n;
  long long x;
  cin >> x;
  while (x--) {
    long long start, endd, nm;
    cin >> start >> endd >> nm;
    long long lower =
        lower_bound(v[nm].begin(), v[nm].end(), start) - v[nm].begin();
    long long upper =
        lower_bound(v[nm].begin(), v[nm].end(), endd) - v[nm].begin();
    if (v[nm][upper] == endd) {
      cout << (upper - lower) + 1 << endl;
    } else {
      cout << upper - lower << endl;
    }
  }
  return 0;
}