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 <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1e18;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("YES");}else{puts("NO");}
const int MX = 51;

// Mod int
const int mod = 1000000007;
struct mint {
  ll x;
  mint():x(0){}
  mint(ll x):x((x%mod+mod)%mod){}
  // mint(ll x):x(x){}
  mint& fix() { x = (x%mod+mod)%mod; return *this;}
  mint& operator+=(const mint& a){ if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a){ (x*=a.x)%=mod; return *this;}
  mint& operator/=(const mint& a){ (x*=a.ex(mod-2).x)%=mod; return *this;}
  mint operator+(const mint& a)const{ return mint(*this) += a;}
  mint operator-(const mint& a)const{ return mint(*this) -= a;}
  mint operator*(const mint& a)const{ return mint(*this) *= a;}
  mint operator/(const mint& a)const{ return mint(*this) /= a;}
  mint ex(ll t) const {
    if(!t) return 1;
    mint res = ex(t/2);
    res *= res;
    return (t&1)?res*x:res;
  }
  bool operator<(const mint& a)const{ return x < a.x;}
  bool operator==(const mint& a)const{ return x == a.x;}
};
mint ex(mint x, ll t) { return x.ex(t);}
istream& operator>>(istream&i,mint&a){i>>a.x;return i;}
ostream& operator<<(ostream&o,const mint&a){o<<a.x;return o;}
typedef vector<mint> vm;
struct comb {
  vm f, g;
  comb(){}
  comb(int mx):f(mx+1),g(mx+1) {
    f[0] = 1;
    rrep(i,mx) f[i] = f[i-1]*i;
    g[mx] = f[mx].ex(mod-2);
    for(int i=mx;i>0;i--) g[i-1] = g[i]*i;
  }
  mint c(int a, int b) {
    if (a < b) return 0;
    return f[a]*g[b]*g[a-b];
  }
} c(55);
//


bitset<55> g[55][55][55][55];
void init(int n) {
  rep(i,n+1)rep(j,i+1) {
    g[i][j][j][0][0] = 1;
    rep(t0,n+1)rep(t1,n+1)rep(k,i+1) if (g[i][j][k][t0][t1]) {
      if (k) g[i][j][k-1][t0+1][t1] = 1;
      if (k<i) g[i][j][k+1][t0][t1+1] = 1;
    }
  }
}

bool vis[55][55][55];
mint dp[55][55][55];
mint dfs(int n, vi s) {
  mint& now = dp[n][s[0]][s[1]];
  if (vis[n][s[0]][s[1]]) return now;
  vis[n][s[0]][s[1]] = true;
  int ss = s[0]+s[1];
  n -= ss;
  if (n < 0) return now = 0;
  if (!n && s == (vi){0,0}) return now = 1;
  rep(s0,ss+1) {
    int s1 = ss-s0;
    rep(t0,n+1)rep(t1,n-t0+1) if (g[ss][s[0]][s0][t0][t1]) {
      now += dfs(n,{t0,t1})*c.c(n+ss,s1)*c.c(n+s0,s0);
    }
  }
  return now;
}


int main() {
  int n;
  scanf("%d",&n);
  init(n);
  mint ans = dfs(n,{1,0});
  cout<<ans<<endl;
  return 0;
}




















