#pragma GCC optimize("O3")
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
using base = complex<double>;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define M 1000000007
#define M2 998244353
#define ll long long
#define pll pair<long, long>
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define REPI(i, a, b) for (ll i = b - 1; i >= a; i--)
#define ff first
#define ss second
#define pb push_back
#define db pop_back
#define mp make_pair
#define mt make_tuple
#define g(a, b) get<a>(b)
#define INF (ll)1e18 + 100
#define vl vector<ll>
#define vi vector<int>
#define vll vector<pair<ll, ll>>
#define vii vector<pair<int, int>>
#define all(v) v.begin(), v.end()
#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))
#define INT_SIZE 32
/*SOME BITMASK KNOWLEDGE
1)x & (x - 1):sets the last one bit of x to zero
power of two exactly when x & (x − 1) = 0.
2)x & -x:sets all the one bits to zero, except last one bit
3)x | (x - 1):inverts all the bits after the last one bit*/
#define endl "\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define o_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define o_setll tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
//member functions :
//1. order_of_key(k) : number of elements strictly lesser than k
//2. find_by_order(k) : k-th element in the set
ll powM(ll x, ll y, ll m)
{
    ll ans = 1, r = 1;
    x %= m;
    while (r > 0 && r <= y)
    {
        if (r & y)
        {
            ans *= x;
            ans %= m;
        }
        r <<= 1;
        x *= x;
        x %= m;
    }
    return ans;
}
ll modI(ll a, ll m)
{
    ll m0 = m, y = 0, x = 1;
    if (m == 1)
        return 0;
    while (a > 1)
    {
        ll q = a / m;
        ll t = m;
        m = a % m;
        a = t;
        t = y;
        y = x - q * y;
        x = t;
    }
    if (x < 0)
        x += m0;
    return x;
}

ll logint(ll x, ll y)
{
    ll ans = 0;
    ll a = 1;
    for (ll i = 0; i <= x; i++)
    {
        if (x < a)
        {
            return ans;
        }
        ans++;
        a *= y;
    }
    return -1;
}
ll extended(ll a, ll b, ll &x, ll &y)
{
    if (a == 0)
    {
        x = 0;
        y = 1;
        return b;
    }
    ll x1, y1;
    ll d = extended(b % a, a, x1, y1);
    x = y1 - (b / a) * x1;
    y = x1;
    return d;
}

// void check(ll i, ll &val) {
//     if (i == 1) {return;}
//     val++;
//     check(__builtin_popcountll(i), val);
// }
// ll dp[1005][1005][2];
ll dp[3005][3005];
void solve() {
    // Let's first check number of ones in the required ans. Then we will find the number of such values.
    // string n;
    // cin >> n;
    // ll k;
    // cin >> k;
    // if (k == 0) {cout << 1; return;}
    // set<ll>s;
    // REP(i, 1, 1001) {
    //     ll val = 1;
    //     check(i, val);
    //     if (val == k) {
    //         s.insert(__builtin_popcountll(i));
    //     }
    // }
    // // cout << *s.begin() << "\n";
    // if (s.empty()) {cout << 0; return;}
    // // Use digit-dp to solve.
    // memset(dp, 0, sizeof(dp));
    // dp[0][0][0] = 1;
    // dp[0][1][1] = 1;
    // REP(i, 0, n.size() - 1) {
    //     REP(j, 0, 1001) {
    //         if (dp[i][j][0]) {
    //             dp[i + 1][j][0] += dp[i][j][0];
    //             dp[i + 1][j + 1][0] += dp[i][j][0];
    //             dp[i + 1][j][0] %= M;
    //             dp[i + 1][j + 1][0] %= M;
    //         }
    //         if (dp[i][j][1]) {
    //             if (n[i + 1] == '0') {
    //                 dp[i + 1][j][1] += dp[i][j][1];
    //                 dp[i + 1][j][1] %= M;
    //             }
    //             else {
    //                 dp[i + 1][j][0] += dp[i][j][1];
    //                 dp[i + 1][j][0] %= M;
    //                 dp[i + 1][j + 1][1] += dp[i][j][1];
    //                 dp[i + 1][j + 1][1] %= M;
    //             }
    //         }
    //     }
    // }
    // ll ans = 0;
    // for (auto y : s) {
    //     ans += (dp[n.size() - 1][y][0] + dp[n.size() - 1][y][1]) % M;
    //     ans %= M;
    // }
    // if (k == 1) {
    //     ans--;
    //     ans += M;
    //     ans %= M;
    // }
    // cout << ans << "\n";
    ll a, b, c, d;
    cin >> a >> b >> c >> d;
    REP(i, 1, a + 1) {
        REP(j, 1, b + 1) {
            dp[i][j] = 1;
        }
    }
    REP(i, a + 1, c + 1) {
        REP(j, 1, b + 1) {
            dp[i][j] = (dp[i - 1][j] * j) % M2;
        }
    }
    REP(j, b + 1, d + 1) {
        REP(i, 1, a + 1) {
            dp[i][j] = (dp[i][j - 1] * i) % M2;
        }
    }
    REP(i, a + 1, c + 1) {
        REP(j, b + 1, d + 1) {
            dp[i][j] = ((dp[i - 1][j] * j) % M2 + (dp[i][j - 1] * i) % M2 - (dp[i - 1][j - 1] * (j - 1) * (i - 1)) % M2 + M2) % M2;
        }
    }
    cout << dp[c][d] % M2 << "\n";
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("input.txt", "w", stdout);
    ll ntc;
    // cin >> ntc;
    ntc = 1;
    REP(tc, 1, ntc + 1)
    {
        // cout << "Case #" << tc << ": ";
        solve();
    }
}

Measured memory_bytes: 74016