#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;

ll fac[200000+1],ifac[200000+1];
 
ll mpow(ll x,ll n){
    ll res = 1;
    while(n != 0){
        if(n&1) res = res*x % mod;
        x = x*x % mod;
        n = n >> 1; 
    }
    return res;
}
ll comb(ll a,ll b){
    if(a==0 && b==0) return 1;
    if(a<b || a<0) return 0;
    ll tmp = ifac[a-b]*ifac[b] % mod;
    return tmp*fac[a] % mod;
}
void calc_fac(){
   fac[0] = 1;
   ifac[0] = 1;
   for(int i = 0;i < 200000;i ++){
       fac[i+1] = fac[i]*(i+1) % mod;
       ifac[i+1] = ifac[i]*mpow(i+1,mod-2) % mod;
   }
}
/*
写像十二相より
玉n個区別箱k個区別しないもとで箱に1個以上入るようなとき
S(n,k) = S(n-1,k-1) + kS(n-1,k) の漸化式より計算
O(nk) だが全ての状態を保存できる
*/
template <typename T> struct Stirling{
    
    vector<vector<T> > S;
    constexpr Stirling(int MAX) noexcept : S(MAX+1, vector<T>(MAX+1, 0)) {
        S[0][0] = 1;
        for (int n = 1; n <= MAX; ++n) {
            for (int k = 1; k <= n; ++k) {
                S[n][k] = (S[n-1][k-1] + S[n-1][k] * k)%mod;
            }
        }
    }
    constexpr T stir(int n, int k) {
        if (n < 0 || k < 0 || n < k) return 0;
        return S[n][k];
    }
};

int main(){
    int n;
    ll ans=0;
    cin >> n;
    Stirling<ll> s(555);
  	calc_fac();
    for(int i = 1;i <= n;i ++){
        for(int x = 1;x <= n;x ++){
            ans = (ans + comb(n,x)*s.stir(x,i)%mod*mpow(i*(i-1),n-x)%mod)%mod;
        }
    }
    cout << ans << endl;
    return 0;
}