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>
#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:210000000")
using namespace std;
const int INF = 2e9 + 9;
const long long INF1 = 8e15 + 9;
const long long MAXN = 3e6 + 7;
const long long MAXN1 = 20;
const long long MAXN2 = 2e7;
const long long MOD = 998244353;
const long long MOD1 = 1e9 + 9;
const long long ALPH = 27;
const long long PW1 = 239;
const long long PW2 = 199;
const long long PW3 = 193;
const long long PW4 = 117;
const long double EPS = 1e-13;
const long long BLOCK = 116;
const long long BLOCK1 = 1 << 9;
void solve();
signed main() {
  srand('a' + 'l' + 'e' + 'x' + 'X' + '5' + '1' + '2');
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  if (0) cin >> q;
  while (q--) solve();
}
vector<long long> prime;
vector<int> vec;
int dp[MAXN], dp1[MAXN];
void rec(int id, long long now, long long r) {
  if (id == prime.size()) return;
  for (long long i = 0, c = 1; now * c <= r; ++i, c *= prime[id]) {
    if (i != 0) vec.emplace_back(now * c);
    rec(id + 1, now * c, r);
  }
}
void solve() {
  for (int i = 2; i <= 100; ++i) {
    bool b = 1;
    for (int j = 2; j * j <= i && b; ++j) {
      b &= (i % j != 0);
    }
    if (b) prime.emplace_back(i);
  }
  int l, r, p;
  cin >> l >> r >> p;
  rec(0, 1, r);
  sort(vec.begin(), vec.end());
  fill(dp, dp + MAXN, INF);
  fill(dp1, dp1 + MAXN, INF);
  dp[0] = 0;
  for (int j = 2; j <= 100; ++j) {
    for (int i = 0, it = 0; i < vec.size(); ++i) {
      if (vec[i] % j == 0) {
        while (vec[it] * j <= vec[i]) ++it;
        dp[i + 1] = min(dp[i + 1], dp[it] + 1);
        dp1[i + 1] = min(dp1[i + 1], dp[i + 1] + j);
      }
    }
  }
  int ans1 = 0;
  for (int i = 0; i < vec.size(); ++i) {
    if (vec[i] >= l && dp1[i + 1] <= p) ++ans1;
  }
  cout << ans1;
}