// 01001100 01001111 01010100 01000001 \\
// \\
// ╦ ╔═╗╔╦╗╔═╗ \\
// ║ ║ ║ ║ ╠═╣ \\
// ╩═╝╚═╝ ╩ ╩ ╩ \\
// \\
// 01001100 01001111 01010100 01000001 \\
#include <bits/stdc++.h>
using namespace std;
#define N 2001
#define nl '\n'
#define ff first
#define ss second
#define add insert
#define ll long long
#define ld long double
#define terminator main
#define pll pair<ll,ll>
#define append push_back
#define pii pair<int,int>
#define all(x) (x).begin(),(x).end()
#define L0TA ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
const int M = 1e9 + 7;
int add(int x, int y){
return (x + y) % M;
}
int mul(int x, int y){
return (1ll * x * y) % M;
}
int dp[N][N][3];
void solve(){
int n, s, e, t;
cin >> n >> s >> e;
t = (s == 1) + (e == 1);
dp[1][1][t] = 1;
for(int i = 2; i <= n; i++){
for(int j = 1; j < i; j++){
if(s == i || e == i){
dp[i][j][t + 1] = add(dp[i][j][t + 1], dp[i - 1][j][t]);
dp[i][j + 1][t + 1] = add(dp[i][j + 1][t + 1], dp[i - 1][j][t]);
}
else{
dp[i][j - 1][t] = add(dp[i][j - 1][t], mul(dp[i - 1][j][t], j - 1));
dp[i][j + 1][t] = add(dp[i][j + 1][t], mul(dp[i - 1][j][t], j + 1 - t));
}
}
t += (s == i || e == i);
}
cout << dp[n][1][2];
}
int terminator(){
L0TA;
solve();
return 0;
}
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