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

template <typename T> struct Combination{

    T mpow(T x,T n){
        T res = 1;
        while(n != 0){
            if(n&1) res = res*x % mod;
            x = x*x % mod;
            n = n >> 1; 
        }
        return res;
    }

    vector<T> fac,ifac;
    constexpr Combination(int max_N) noexcept : fac(max_N,1),ifac(max_N,1){
        fac[0] = 1;
        ifac[0] = 1;
        for(int i = 0;i < max_N;i ++){
            fac[i+1] = fac[i]*(i+1) % mod;
            ifac[i+1] = ifac[i]*mpow(i+1,mod-2) % mod;
        }
    }

    constexpr T comb(T a,T b){
        if(a==0 && b==0) return 1;
        if(a<b || a<0) return 0;
        T tmp = ifac[a-b]*ifac[b] % mod;
        return tmp*fac[a] % mod;
    }
    constexpr T perm(T a,T b){
        if(a==0 && b==0) return 1;
        if(a<b || a<0) return 0;
        return ifac[a-b]*fac[a] % 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 (ll 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);
    Combination<ll> c(555);
    for(int i = 1;i <= n;i ++){
        for(int x = i;x <= n;x ++){
            ans = (ans + (c.comb(n,x)%mod)*(s.stir(x,i)%mod)*(c.mpow(i*(i-1),n-x)%mod))%mod;
        }
    }
    cout << ans << endl;
    return 0;
}