#include <atcoder/all>
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
#define rep(i, n) REP(i, 0, n)
#define REP(i, s, e) for (int i = (s); i < (int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for (int i = (int)(s - 1); i >= (int)(e); i--)
#define all(r) r.begin(), r.end()
#define rall(r) r.rbegin(), r.rend()

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;

template <typename T, typename U>
T chmax(T& a, const U& b) {
    if (a >= b) return false;
    a = b;
    return true;
}
template <typename T, typename U>
T chmin(T& a, const U& b) {
    if (a <= b) return false;
    a = b;
    return true;
}

void yes_no(bool f, string yes = "Yes", string no = "No") { cout << (f ? yes : no) << "\n"; }

void solve() {
    int n;
    cin >> n;
    if (n <= 2) {
        cout << n << "\n";
        return;
    }
    using mint = modint1000000007;
    vector dp(n + 1, vector(n + 1, vector(2, vector(2, mint(0)))));
    dp[2][0][0][0] = 2;
    REP(i, 1, n)
    rep(ng, i) {
        {
            mint tmp = dp[i][ng][1][1];
            if (ng > 0) dp[i + 1][ng - 1][0][0] += tmp;
            dp[i + 1][ng][1][1] += tmp;
            dp[i + 1][ng + 1][1][1] += tmp;
            if (ng > 0) dp[i + 1][ng - 1][1][0] += tmp * (ng - 2);
            dp[i + 1][ng][1][0] += tmp * (i - ng);
        }
        {
            mint tmp = dp[i][ng][1][0];
            dp[i + 1][ng][0][1] += tmp;
            dp[i + 1][ng + 1][0][1] += tmp;
            if (ng > 0) dp[i + 1][ng - 1][0][0] += tmp * (ng - 1);
            dp[i + 1][ng][0][0] += tmp * (i - ng);
        }
        {
            mint tmp = dp[i][ng][0][1];
            if (ng > 0) dp[i + 1][ng - 1][0][0] += tmp;
            dp[i + 1][ng + 1][1][1] += tmp * 2;
            if (ng > 0) dp[i + 1][ng - 1][1][0] += tmp * (ng - 1);
            dp[i + 1][ng][1][0] += tmp * (i - ng - 1);
        }
        {
            mint tmp = dp[i][ng][0][0];
            dp[i + 1][ng + 1][0][1] += tmp * 2;
            if (ng > 0) dp[i + 1][ng - 1][0][0] += tmp * ng;
            dp[i + 1][ng][0][0] += tmp * (i - ng - 1);
        }
    }
    mint ans = dp[n][0][0][0];
    cout << ans.val() << "\n";
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int t = 1;
    // cin >> t;
    rep(ti, t) solve();
    return 0;
}