#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,j,n) for(ll i=j;i<(ll)(n);i++)
#define rrep(i,j,n)  for(ll i=j;i>=n;i--)
#define all(x) (x).begin(),(x).end()
#define m0(x) memset(x,0,sizeof(x))
#define pb push_back
#define mp make_pair
#define perm(c) sort(all(c)); for(bool c##p=1;c##p;c##p=next_permutation(all(c)))
#define UNIQUE(v) sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end())

using namespace std;
using namespace atcoder;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
template<class T> bool chmax(T &a, const T &b){if(a<b) {a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b){if(a>b) {a=b;return 1;}return 0;}
const ll LINF = 1LL << 60LL;
const int IINF = (1 << 30) - 1;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

using mint = modint1000000007;
mint dp[1100][1100][2][2] = {0};
void solve(){
    ll n; cin >> n;
    dp[1][0][0][0] = 1;
    dp[2][0][0][0] = 2;
    rep(i,2,n){
        rep(j,0,i){
            if(j == 0){
                // i+1 i-1 or i-1 i+1
                dp[i+1][j+1][0][1] += 2*dp[i][j][0][0];
                // others
                dp[i+1][j][0][0] += (i-1)*dp[i][j][0][0];
            }
            else{
                // i+1 i-1 or i-1 i+1
                dp[i+1][j+1][0][1] += 2*dp[i][j][0][0];
                // x i+1 x+2
                dp[i+1][j-1][0][0] += j*dp[i][j][0][0];
                // others
                dp[i+1][j][0][0]   += (i-j-1)*dp[i][j][0][0];

                // i-2 i+1 i
                dp[i+1][j-1][0][0] += dp[i][j][0][1];
                // i+1 i-1 or i-1 i+1
                dp[i+1][j+1][1][1] += 2*dp[i][j][0][1];
                // x i+1 x+2
                dp[i+1][j-1][1][0] += (j-1)*dp[i][j][0][1];
                //others
                dp[i+1][j][1][0] += (i-j-1)*dp[i][j][0][1];

                // i+1 i-1 or i-1 i+1
                dp[i+1][j+1][1][1] += dp[i][j][1][1];
                // i-2 i+1 i
                dp[i+1][j-1][0][0] += dp[i][j][1][1];
                // x i+1 x+2
                dp[i+1][j-1][1][0] += (j-2)*dp[i][j][1][1];
                // x-3 i+1 x-1
                dp[i+1][j][1][1] += dp[i][j][1][1];
                // others
                dp[i+1][j][1][0] += (i-j)*dp[i][j][1][1];


                dp[i+1][j+1][0][1] += dp[i][j][1][0];
                dp[i+1][j-1][0][0] += (j-1)*dp[i][j][1][0];
                // others
                dp[i+1][j][0][1]   += dp[i][j][1][0];
                dp[i+1][j][0][0]   += (i-j)*dp[i][j][1][0];
            }
        }
    }
    cout << dp[n][0][0][0].val() << endl;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    ll T;
    T = 1;
    /*
       cin >> T;
       */
    while(T--) solve();
}