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 <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXN = 50;
const int MOD = 1000000007;

int n;
int dp[MAXN + 1][2 * MAXN + 1]; // dp[i][MAXN+j] = # ways to choose first i items such we next need to chose items of value +-2^x and the sum so far is 1+j*2^x

int inv[MAXN + 1];
int fac[MAXN + 1];
int ifac[MAXN + 1];
int C(int n, int k) { if (k<0 || k>n) return 0; return (ll)fac[n] * ifac[k] % MOD*ifac[n - k] % MOD; }

void inc(int &a, int b) { if ((a += b) >= MOD) a -= MOD; }
void run() {
	inv[1] = 1; FORE(i, 2, MAXN) inv[i] = (ll)(MOD - MOD / i)*inv[MOD%i] % MOD;
	fac[0] = 1; FORE(i, 1, MAXN) fac[i] = (ll)fac[i - 1] * i%MOD;
	ifac[0] = 1; FORE(i, 1, MAXN) ifac[i] = (ll)ifac[i - 1] * inv[i] % MOD;

	scanf("%d", &n);
	memset(dp, 0, sizeof(dp)); dp[1][MAXN] = n; dp[1][MAXN - 1] = n;
	REP(i, n) FORE(j, -n, +n) if (dp[i][MAXN + j] != 0) {
		//printf("%d %d: %d\n", i, j, dp[i][MAXN + j]);
		FORE(k, max(1, abs(j)), n - i) if((j+k)%2==0) REPE(l, k) {
			int ni = i + k, nj = (j + k) / 2 - l; if (nj<-n || nj>+n) continue;
			int ways = (ll)dp[i][MAXN + j] * C(n - i, k) % MOD*C(k, l) % MOD;
			//printf("-> %d %d: %d\n", ni, nj, ways);
			inc(dp[ni][MAXN + nj], ways);
		}
	}
	printf("%d\n", dp[n][MAXN + 0]);
}

int main() {
	run();
	return 0;
}
