Problem: Because of budget cuts one IT company established new non-financial reward system instead of bonuses.

Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.

Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.

In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.

One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.

Output

Output one integer — the amount of ways to place the pennants on n tables.

Examples

Input

2


Output

24

Solution (in JAVA):

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

public class Main {
	
	static PrintWriter out;
	
	static double EPS = 1e-9;
	static int INF = (int) 1e9;
	static long LINF = (long) 1e18;
	static int MOD = (int) 1e9 + 7;
	static double PI = 2 * Math.acos(0);
	
	static BigInteger Binomial(int n, int k) {
		BigInteger res = BigInteger.ONE;
		for (int i = 0; i < k; i++) {
			res = res.multiply(BigInteger.valueOf(n - i));
		}
		for (int i = 0; i < k; i++) {
			res = res.divide(BigInteger.valueOf(i + 1));
		}
		return res;
	}
	
	static void solve() {
		int n = Input.nextInt();
		BigInteger ans = Binomial(5 + n - 1, n - 1);
		ans = ans.multiply(Binomial(3 + n - 1, n - 1));
		out.print(ans);
	}
	
	public static void main(String[] args) {
		out = new PrintWriter(System.out);
//		long stime = System.currentTimeMillis();
		solve();
//		out.println("\nTime elapsed: " + (System.currentTimeMillis() - stime) + "ms");
		out.close();
	}
	
	static class SegmentTree {
		int maxn = 100010;
		long[] st = new long[4 * maxn];
		void update(int node, int i, int L, int R, long val) {
			if (i < L || i > R) return;
			if (L == R) {st[node] += val; return;}
			update(node << 1, i, L, (L + R) >> 1, val);
			update((node << 1) + 1, i, ((L + R) >> 1) + 1, R, val);
			st[node] = st[node << 1] + st[(node << 1) + 1];
		}
		long query(int node, int l, int r, int L, int R) {
			if (l > R || r < L) return 0;
			if (l <= L && r >= R) return st[node];
			return query(node << 1, l, r, L, (L + R) >> 1) + query((node << 1) + 1, l, r, ((L + R) >> 1) + 1, R);
		}
	}
	
	static class Matrix {
		int maxn = 2;
		int[][] x = new int[maxn][maxn];
		Matrix mul(Matrix A) {
			Matrix res = new Matrix();
			for (int i = 0; i < maxn; i++) {
				for (int k = 0; k < maxn; k++) {
					for (int j = 0; j < maxn; j++) {
						res.x[i][j] = (int) ((res.x[i][j] + (long) x[i][k] * A.x[k][j]) % MOD);
					}
				}
			}
			return res;
		}
		Matrix pow(long k) {
			Matrix res = this, T = this; k--;
			while (k > 0) {
				if ((k & 1) > 0) {
					res = res.mul(T);
				}
				T = T.mul(T); k >>= 1;
			}
			return res;
		}
	}
	
	static class Pair implements Comparable<Pair> {
		long fi, se;
		Pair(long fi, long se) {
			this.fi = fi;
			this.se = se;
		}
		public int compareTo(Pair oth) {
			if (fi > oth.fi) return +1;
			if (fi < oth.fi) return -1;
			if (se > oth.se) return +1;
			if (se < oth.se) return -1;
			return 0;
		}
	}
	
	static long gcd(long a, long b) {
		long r;
		while (b != 0) {
			r = a % b;
			a = b;
			b = r;
		}
		return a;
	}
	
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}
	
	static long fpow(long n, long k, int p) {
		long r = 1;
		for (; k > 0; k >>= 1) {
			if (k % 2 == 1) {
				r = r * n % p;
			}
			n = n * n % p;
		}
		return r;
	}
	
	static BigInteger FloorOfNthRoot(BigInteger x, int n) {
	    int sign = x.signum();
	    if (n <= 0 || sign < 0) {
	        throw new IllegalArgumentException();
	    }
	    if (sign == 0) {
	        return BigInteger.ZERO;
	    }
	    if (n == 1) {
	        return x;
	    }
	    BigInteger a;
	    BigInteger bigN = BigInteger.valueOf(n);
	    BigInteger bigNMinusOne = BigInteger.valueOf(n - 1);
	    BigInteger b = BigInteger.ZERO.setBit(1 + x.bitLength() / n);
	    do {
	        a = b;
	        b = a.multiply(bigNMinusOne).add(x.divide(a.pow(n - 1))).divide(bigN);
	    }
	    while (b.compareTo(a) == -1);
	    return a;
	}
	
	static class Input {
		static StringTokenizer token = null;
		static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//		static BufferedReader in;
//		static {
//			try {
//				in = new BufferedReader(new FileReader("in.txt"));
//			} catch (IOException e) {
//			}
//		}
		static int nextInt() {
			return Integer.parseInt(nextString());
		}
		static long nextLong() {
			return Long.parseLong(nextString());
		}
		static double nextDouble() {
			return Double.parseDouble(nextString());
		}
		static String nextString() {
			try {
				while (token == null || !token.hasMoreTokens()) {
					token = new StringTokenizer(in.readLine());
				}
			} catch (IOException e) {
			}
			return token.nextToken();
		}
	}
}