There are n boys and m girls attending a theatre club. To set a play "The Big Bang Theory", they need to choose a group containing exactly t actors containing no less than 4 boys and no less than one girl. How many ways are there to choose a group? Of course, the variants that only differ in the composition of the troupe are considered different.

Perform all calculations in the 64-bit type: long long for С/С++, int64 for Delphi and long for Java.

Input

The only line of the input data contains three integers n, m, t (4 ≤ n ≤ 30, 1 ≤ m ≤ 30, 5 ≤ t ≤ n + m).

Output

Find the required number of ways.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.

Examples

Input

5 2 5


Output

10


Input

4 3 5


Output

3
//package round95;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {
	static BufferedReader bf = new BufferedReader(new InputStreamReader(
			System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);

	static String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String s = bf.readLine();
			if (s == null)
				return null;
			st = new StringTokenizer(s);
		}

		return st.nextToken();
	}

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

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

	static String nextStr() throws IOException {
		return nextToken();
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	
	static void f(int n, long c[][]) {
		for (int i = 0; i <= n; i++) {
			c[0][i] = 1;
			for (int j = 1; j <= i; j++) {
				c[j][i] = c[j][i - 1] + c[j - 1][i - 1];
			}
		}
	}

	public static void main(String[] args) throws IOException {
		int n = nextInt();
		int m = nextInt();
		int t = nextInt();
		
		long c1[][] = new long[n + 1][n + 1];
		long c2[][] = new long[m + 1][m + 1];
		f(n, c1);
		f(m, c2);
		
		long res = 0;
		for (int i = 4; i < t && i <= n; i++) {
			if (t - i <= m) {
				res += c1[i][n] * c2[t - i][m];
			}
		}
		
		out.println(res);
		out.flush();
	}
}
