Problem: Maxim has got a calculator. The calculator has two integer cells. Initially, the first cell contains number 1, and the second cell contains number 0. In one move you can perform one of the following operations:

  1. Let's assume that at the current time the first cell contains number a, and the second cell contains number b. Write to the second cell number b + 1; 
  2. Let's assume that at the current time the first cell contains number a, and the second cell contains number b. Write to the first cell number a·b. 



Maxim is wondering, how many integers x (l ≤ x ≤ r) are there, such that we can write the number x to the first cell of the calculator, having performed at most p moves.

Input

The first line contains three integers: l, r, p (2 ≤ l ≤ r ≤ 109, 1 ≤ p ≤ 100). 

The numbers in the line are separated by single spaces.

Output

In a single line print a single integer — the answer to the problem.

Examples

Input

2 10 3


Output

1


Input

2 111 100


Output

106


Input

2 111 11


Output

47

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T in() {
  char ch;
  T n = 0;
  bool ng = false;
  while (1) {
    ch = getchar();
    if (ch == '-') {
      ng = true;
      ch = getchar();
      break;
    }
    if (ch >= '0' && ch <= '9') break;
  }
  while (1) {
    n = n * 10 + (ch - '0');
    ch = getchar();
    if (ch < '0' || ch > '9') break;
  }
  return (ng ? -n : n);
}
template <typename T>
inline T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return (POW(B, P / 2) * POW(B, P / 2));
}
template <typename T>
inline T Gcd(T a, T b) {
  if (a < 0) return Gcd(-a, b);
  if (b < 0) return Gcd(a, -b);
  return (b == 0) ? a : Gcd(b, a % b);
}
template <typename T>
inline T Lcm(T a, T b) {
  if (a < 0) return Lcm(-a, b);
  if (b < 0) return Lcm(a, -b);
  return a * (b / Gcd(a, b));
}
long long Bigmod(long long base, long long power, long long MOD) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return ret;
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
template <typename T>
long long int isLeft(T a, T b, T c) {
  return (a.x - b.x) * (b.y - c.y) - (b.x - c.x) * (a.y - b.y);
}
long long ModInverse(long long number, long long MOD) {
  return Bigmod(number, MOD - 2, MOD);
}
bool isConst(char ch) {
  if (isalpha(ch) && !isVowel(ch)) return true;
  return false;
}
int toInt(string s) {
  int sm;
  stringstream second(s);
  second >> sm;
  return sm;
}
bool prime[109];
vector<int> prm;
void seive() {
  prime[1] = true;
  int s = 100;
  for (int i = 2; i <= s; i++) {
    if (!prime[i]) {
      for (int j = 2 * i; j <= 100; j += i) {
        prime[j] = true;
      }
    }
  }
  for (int i = 2; i < 100; i++)
    if (prime[i] == false) prm.push_back(i);
}
int a[3000009];
int Ln;
vector<long long int> v[102];
int Pln;
void Gen(int id) {
  if (id == Pln) return;
  int now = Ln;
  for (int i = 0; i < now; i++) {
    for (int j = 0; j < v[id].size(); j++) {
      long long int x = (v[id][j] * a[i]);
      if (x <= 1e9) a[Ln++] = x;
    }
  }
  Gen(id + 1);
}
int dp[3000009];
bool Ok[3000009];
int main() {
  seive();
  Pln = prm.size();
  for (int i = 0; i < Pln; i++) {
    long long int nw = prm[i];
    while (nw <= 1e9) {
      v[i].push_back(nw);
      nw *= prm[i];
    }
  }
  a[Ln++] = 1;
  Gen(0);
  sort(a, a + Ln);
  for (int i = 0; i < Ln; i++) dp[i] = 2000000;
  long long int l, r, p;
  l = in<long long int>(), r = in<long long int>(), p = in<long long int>();
  dp[0] = 0;
  Ok[0] = 1;
  for (int i = 2; i <= p; i++) {
    int pos = 0;
    for (int j = 0; j < Ln; j++) {
      while (pos < Ln && a[pos] < a[j] * i) pos++;
      if (a[pos] == (a[j] * i)) {
        dp[pos] = min(dp[pos], dp[j] + 1);
        if (dp[pos] + i <= p) Ok[pos] = 1;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < Ln; i++) {
    if (Ok[i] && (a[i] >= l && a[i] <= r)) ans++;
  }
  printf("%d\n", ans);
  return 0;
}