Let X = 10^{100}. Inaba has N checker pieces on the number line, where the i-th checker piece is at coordinate X^{i} for all 1 \leq i \leq N.

Every second, Inaba chooses two checker pieces, A and B, and move A to the symmetric point of its current position with respect to B. After that, B is removed. (It is possible that A and B occupy the same position, and it is also possible for A to occupy the same position as another checker piece after the move).

After N - 1 seconds, only one checker piece will remain. Find the number of distinct possible positions of that checker piece, modulo 10^{9} + 7.

Constraints

* 1 \leq N \leq 50
* N is an integer.

Input

Input is given from Standard Input in the following format:


N


Output

Print the number of distinct possible positions of the final checker piece, modulo 10^{9} + 7.

Examples

Input

3


Output

12


Input

4


Output

84


Input

22


Output

487772376
#include <iostream>
using namespace std;

long long N, dp[59][59][59], nr[59][59], mod = 1000000007;

void init() {
	for (int i = 0; i < 59; i++) {
		for (int j = 0; j < 59; j++) {
			if (i == 0 || j == 0) nr[i][j] = 1;
			else nr[i][j] = (nr[i - 1][j] + nr[i][j - 1]) % mod;
		}
	}
}

long long ncr(long long n, long long r) {
	return nr[n - r][r];
}

long long solve(int pos, int ca, int cb) {
	if (pos == N) {
		if (ca == 0 && cb == 0) return 1;
		return 0;
	}
	if (dp[pos][ca][cb] >= 0) return dp[pos][ca][cb];

	long long sum = 0;
	for (int i = 0; i <= N - pos; i++) {
		for (int j = 0; j <= N - pos; j++) {
			if ((i + j) < (ca + cb)) continue;
			if ((i + j) % 2 != (ca + cb) % 2) continue;
			if (pos + i + j > N || i + j == 0) continue;

			long long la = 0, lb = 0;
			if ((ca - cb) > (i - j)) la = ((ca - cb) - (i - j)) / 2; 
			else lb = ((i - j) - (ca - cb)) / 2;

			long long P1 = ncr(pos + i, i);
			long long P2 = ncr(pos + i + j, j);
			long long P3 = solve(pos + i + j, la, lb);
			sum += (P1*P2%mod)*P3%mod; sum %= mod;
		}
	}
	dp[pos][ca][cb] = sum;
	return sum;
}

int main() {
	cin >> N; init();
	for (int i = 0; i <= N; i++) { for (int j = 0; j <= N; j++) { for (int k = 0; k <= N; k++) dp[i][j][k] = -1; } }

	long long A = solve(1, 0, 0);
	long long B = solve(1, 1, 0);
	cout << (A + B) % mod << endl;
	return 0;
}