// Approach 1: Using Dynamic Programming

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <chrono>
#include <climits>
#include <cmath>
#include <cstdio>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;

template <class T1, class T2>
ostream& operator << (ostream& out, const pair <T1, T2> p)
{
    out << '(' << p.first << ',' << p.second << ')';
    return out;
}

template <class T1, class T2>
istream& operator >> (istream& in, pair<T1, T2> &p)
{
    in >> p.first >> p.second;
    return in;
}

template <class T>
istream& operator >> (istream& in, vector<T> &v)
{
    for (T &x : v)
        in >> x;
    return in;
}

template <class T>
ostream& operator << (ostream& out, const vector<vector<T>> &v)
{
    for (const vector<T> &x : v)
        out << x << '\n';
    return out;
}

template <class T>
ostream& operator << (ostream& out, const vector<T> &v)
{
    for (const T &x : v)
        out << x << ' ';
    return out;
}

long long gcd (long long a, long long b)
{
    if (b > a)
        swap(a, b);
    return (b ? gcd(b, a % b) : a);
}

using ll   = long long;
using pii  = pair<int, int>;
using pll  = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi   = vector<int>;
using vl   = vector<long long>;
using vvi  = vector<vector<int>>;
using vvl  = vector<vector<long long>>;

#define F          first
#define S          second
#define First      first.first
#define Second     first.second
#define Third      second
#define mp         make_pair
#define rep(i,a,b) for (int i = (a); i < (b); i++)
#define per(i,b,a) for (int i = (b); i > (a); i--)
#define all(x)     x.begin(), x.end()
#define ret(x)     return cout << x, 0;
#define throwex    throw runtime_error ("Found the error.");

const int h = 1000000007;

namespace modexp
{
    const int mod = 1000000007;

    constexpr long long p (long long x, long long y)
    {
        long long ans = 1;
        while(y)
        {
            if(y & 1)
                (ans *= x) %= mod;
            (x *= x) %= mod;
            y /= 2;
        }
        return ans;
    }

    constexpr long long inv (const long long x)
    {
        return p (x, mod - 2);
    }
}
using modexp::p;
using modexp::inv;

signed main()
{
    ios::sync_with_stdio(false);
    #ifdef ONLINE_JUDGE
    cin.tie(nullptr);
    cout.tie(nullptr);
    cerr.setstate(ios::failbit);
    #endif

    int n;
    cin >> n;
    vi p(n);
    cin >> p;
    rep(i,0,n)
        if(p[i] != -1)
            p[i]--;
    vvi M(n);
    rep(i,0,n)
    {
        if(p[i] != -1)
        {
            M[i].push_back(p[i]);
            M[p[i]].push_back(i);
        }
    }
    vi vis(n);
    vi trees;
    int num_cycles = 0;
    rep(i,0,n)
    {
        if(vis[i])
            continue;
        stack<pii> st;
        vis[i] = true;
        st.push({-1, i});
        bool found_cycle = false;
        int cnt = 0;
        while(!st.empty())
        {
            cnt++;
            int prev = st.top().F;
            int cur = st.top().S;
            st.pop();
            for(int x : M[cur])
            {
                if(x == prev)
                    continue;
                if(vis[x])
                {
                    found_cycle = true;
                    continue;
                }
                vis[x] = true;
                st.push({cur, x});
            }
        }
        if(!found_cycle)
            trees.push_back(cnt);
        else
            num_cycles++;
    }
    cerr << trees << '\n';
    vl fact(trees.size()+1);
    fact[0] = 1;
    rep(i,1,trees.size()+1)
        fact[i] = fact[i-1] * i % h;
    ll ans = ::p(n-1, trees.size()) * (n - num_cycles) % h;
    if(trees.empty())
        ret(ans);
    vvl dp(trees.size(), vl(trees.size()));
    dp[0][0] = trees[0];
    rep(i,1,trees.size())
    {
        dp[i][0] = dp[i-1][0] + trees[i];
        rep(j,1,trees.size())
            dp[i][j] = (dp[i-1][j] + dp[i-1][j-1] * trees[i] % h) % h;
    }
    cerr << dp;
    rep(i,1,trees.size())
        ((ans -= dp.back()[i] * fact[i] % h * ::p(n-1, trees.size()-i-1) % h) += h) %= h;
    rep(i,0,trees.size())
        ((ans -= (trees[i]-1) * ::p(n-1, trees.size()-1) % h) += h) %= h;
    cout << ans;
}


Measured memory_bytes: 199012