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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class E {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	static final int[] PS = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
			43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 };

	int l, r, p;
	List<Integer> goodAL = new ArrayList<>();

	static int[] toIntArray(List<Integer> a) {
		int[] ret = new int[a.size()];
		int i = 0;
		for (Integer x : a) {
			ret[i++] = x;
		}
		return ret;
	}

	void dfs(int x, int i) {
		if (i == PS.length) {
			return;
		}
		dfs(x, i + 1);
		if (x <= r / PS[i]) {
			int next = x * PS[i];
			goodAL.add(next);
			dfs(next, i);
		}
	}

	void solve() throws IOException {
		l = nextInt();
		r = nextInt();
		p = nextInt();
		goodAL.add(1);
		dfs(1, 0);
		int[] good = toIntArray(goodAL);
		Arrays.sort(good);

		int n = good.length;

		int[] dp = new int[n];
		int oo = Integer.MAX_VALUE / 3;
		Arrays.fill(dp, oo);
		dp[0] = 0;
		for (int x = 2; x <= p; x++) {
			for (int i = 0, j = 0; i < n; i++) {
				long needL = (long) good[i] * x;
				if (needL > r) {
					break;
				}
				while (j < n && good[j] < needL) {
					j++;
				}
				if (j == n || good[j] != needL) {
					continue;
				}
				if (dp[i] + x < p) {
					int a = dp[j];
					int b = dp[i] + 1;
					dp[j] = (a <= b) ? a : b;
				}
			}
		}

		int ans = 0;

		for (int i = 0; i < n; i++) {
			if (good[i] >= l && dp[i] != oo) {
				ans++;
			}
		}

		out.println(ans);
	}

	E() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new E();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}