#include <bits/stdc++.h>
using namespace std;
#define int long long
// has to be DP:
// most probably dp[i][j] for n^2 solution
// buckets modulo M: each bucket is 0 to M-1, the positions of the stones must not be in the same bucket
// pair up 2n positions into n pairs: 1 from 1 bucket, other form other bucket
// we can match up positions, then multiply by n! (for each different possible config )
// and 2^n for switching stone color
// should be dp:
// from using positions from 0 to i
// j is amount of stuff left over from previous buckets: we can get at most j + 1 things from the current bucket to pair off
// transition may take too long
const int MOD = 1e9 + 7;
const int MAXN = 4005;
int powmod(int a, int b){
int res = 1;
while(b){
if(b & 1){
res = res * a % MOD;
}
a = a*a % MOD;
b >>= 1;
}
return res;
}
int factorial[MAXN], invfact[MAXN];
int C(int n, int k){
if(k < 0 || k > n){
return 0;
}
return (factorial[n] * ((invfact[k] * invfact[n-k])%MOD))%MOD;
}
int32_t main(){
int n, m;
cin >> n >> m;
int tot = 2*n;
vector<int> mods(m, 0);
for(int i = 1; i <= tot; i++){
mods[i % m]++;
}
sort(mods.begin(), mods.end());
factorial[0] = 1;
for(int i = 1; i <= tot; i++){
factorial[i] = factorial[i-1] * i % MOD;
}
invfact[tot] =powmod(factorial[tot], MOD - 2);
for(int i = tot; i > 0; i--){
invfact[i-1] = invfact[i] * i % MOD;
}
vector<int> dp(tot+1, 0), ndp(tot + 1, 0);
dp[0] = 1;
int processed = 0;
for(int i = 0; i < m; i++){
int bucket = mods[i];
if(bucket == 0){
continue;
}
fill(ndp.begin(), ndp.end(), 0);
for(int j = 0; j <= processed; j++){
if(dp[j] == 0){
continue;
}
for(int k = 0; k <= min(k, bucket); k++){
int nj = j + bucket - 2 * k;
int remaining = tot - (processed + bucket);
if(nj > remaining){
continue; // doesnt work
}
int ways = dp[j];
ways *= C(j, k);
ways %= MOD;
ways *= C(bucket, k);
ways %= MOD;
ways *= factorial[k];
ways %= MOD;
ndp[nj] += ways;
ndp[nj] %= MOD;
}
}
processed += bucket;
dp.swap(ndp);
}
int ans = dp[0];
ans *= factorial[n];
ans %= MOD;
ans *= powmod(2, n);
ans %= MOD;
cout << ans << '\n';
}
