#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define sz(x) (int) x.size()
#define pb push_back
#define _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int ll
//#define gato
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ii = pair<int,int>;
using iii = tuple<int,int,int>;
const int inf = 2e9+1;
const int mod = 1e9+7;
const int maxn = 3e5+100;
template<typename X, typename Y> bool ckmin(X& x, const Y& y) { return (y < x) ? (x=y,1):0; }
template<typename X, typename Y> bool ckmax(X& x, const Y& y) { return (x < y) ? (x=y,1):0; }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) {
uniform_int_distribution<int> uid(l, r);
return uid(rng);
}
int dp[2005], ncr[2005][2005];
int n3(int n, int a, int b) {
if (n == 2) return 1;
if (a > b) swap(a, b);
vector f(n+1, vector(n+1, vector(2, 0LL)));
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i; ++j) {
if (i == 1) {
f[i][i][1] = 1;
continue;
}
// f[i][j][0] = \sum f[i-1][k][1], k < j
for (int k = 1; k < j; ++k) f[i][j][0] += f[i-1][k][1];
f[i][j][0] %= mod;
for (int k = j+1; k <= i; ++k) f[i][j][1] += f[i-1][k-1][0];
f[i][j][1] %= mod;
}
}
if (b == n) return (f[n-1][a][0] + f[n-1][a][1]) % mod;
int ans = 0;
for (int s1 = 1; s1 <= n-2; ++s1) {
int s2 = n-1-s1;
for (int i1 = 1; i1 <= min(a, s1); ++i1) {
for (int i2 = 0; i1 + i2 <= s1 and i2 < b-a; ++i2) {
int j = b-i1-i2;
if (j > s2) continue;
int ways = ncr[a-1][i1-1] * ncr[b-a-1][i2] % mod * ncr[n-1-b][s2-j] % mod;
if (n & 1) ways = ways * ((f[s1][i1][0] * f[s2][j][0]) % mod + (f[s1][i1][1] * f[s2][j][1]) % mod) % mod;
else ways = ways * ((f[s1][i1][1] * f[s2][j][0]) % mod + (f[s1][i1][0] * f[s2][j][1]) % mod) % mod;
ans = (ans + ways) % mod;
}
}
}
return ans;
}
void solve() {
int n, a, b; cin >> n >> a >> b;
// dp[direcao][round][posicao]
// quero fazer uma inclusao exclusao mesmo? acho que nao!
dp[1] = 1;
for (int i = 2; i <= n; ++i) {
dp[i] = max(1LL, n-i) * dp[i-1] % mod;
for (int j = 3; j < i; ++j) {
int ways = ncr[n-i+j-(i!=n)][j] * dp[i-j] % mod;
if (j & 1) dp[i] = (dp[i] + ways) % mod;
else dp[i] = (dp[i] + mod - ways) % mod;
}
int livres = n - a - (b > a);
if (i-1 <= livres and i != 2) {
if (i & 1) dp[i] = (dp[i] + mod - ncr[livres][i-1]) % mod;
else dp[i] = (dp[i] + ncr[livres][i-1]) % mod;
}
}
}
int brute(int n, int a, int b) {
vector<int> p(n);
iota(all(p), 1);
int ans = 0;
do {
bool ok = (p[0] == a and p[n-1] == b);
for (int i = 2; i < n; ++i) ok &= (p[i] < p[i-1]) != (p[i-1] < p[i-2]);
if (ok) ans++;
} while(next_permutation(all(p)));
return ans;
}
int32_t main() {_
for (int i = 0; i <= 2000; ++i) for (int j = 0; j <= i; ++j) {
if (j == 0 or j == i) ncr[i][j] = 1;
else ncr[i][j] = (ncr[i-1][j] + ncr[i-1][j-1]) % mod;
}
#ifndef gato
int n, a, b; cin >> n >> a >> b;
//cout << brute(n, a, b) << endl;
cout << n3(n, a, b) << endl;
#else
int t = 1;
while (true) {
int n = rnd(2, 10), a = rnd(1, n-1), b = rnd(a+1, n);
int my = n3(n, a, b), ans = brute(n, a, b);
if (my != ans) {
cout << "Wrong answer on test " << t << endl;
cout << n << " " << a << " " << b << endl;
cout << "Solve: " << my << endl;
cout << "Brute: " << ans << endl;
exit(0);
}
cout << "Accepted on test " << t++ << endl;
}
#endif
}
