// P9896 [ICPC2018 Qingdao R] Sub-cycle Graph 
#include <cstdio>
#include <iostream>
#define ll long long
#define rep(i, s, t) for(int i=s; i<=t; ++i)
#define debug(x) cerr<<#x<<":"<<x<<endl;
const int N=100010, mod=1e9+7;
using namespace std;
char buf[1<<23], *p1=buf, *p2=buf;
#define gc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
inline int read()
{
    int x=0, f=1; char c=gc();
    while(c<'0' || c>'9') c=='-' && (f=-1), c=gc();
    while('0'<=c && c<='9') x=(x<<3)+(x<<1)+c-'0', c=gc();
    return x*f;
}

int T, n, m; ll res, tmp;
ll fac[N], inv[N], inv2[N], Inv2;
ll qp(ll x, ll y=mod-2)
{
    ll res=1;
    for(; y; y>>=1, x=x*x%mod) if(y&1) res=res*x%mod;
    return res;
}
ll C(int n, int m)
{
    if(n<m) return 0;
    return fac[n]*inv[n-m]%mod*inv[m]%mod;
}
void solve()
{
    n=read(), m=read(); res=-1;
    if(!m) res=1;
    if(m>n) res=0;
    if(m==n) res=fac[n-1]*Inv2%mod;
    if(~res) {printf("%lld\n", res); return;}
    res=0;
    rep(x, 1, n/2)
    {
        int d1=x+x, d2=(m+m-d1)>>1, d3=n-d1-d2;
        tmp=C(n, d1)*C(n-d1, d2)%mod;
        tmp=tmp*C(d1, x)%mod*fac[x]%mod*inv2[x]%mod;
        tmp=tmp*fac[d2]%mod*C(d2+x-1, x-1)%mod;
        res=(res+tmp)%mod;
    }
    printf("%lld\n", res);
}

int main()
{
#ifdef Jerrywang
    freopen("../in.txt", "r", stdin);
#endif
    fac[0]=1; rep(i, 1, N-1) fac[i]=fac[i-1]*i%mod;
    inv[N-1]=qp(fac[N-1]), Inv2=qp(2);
    for(int i=N-1; i; i--) inv[i-1]=inv[i]*i%mod;
    inv2[0]=1; rep(i, 1, N-1) inv2[i]=inv2[i-1]*Inv2%mod;
    T=read(); while(T--) solve();
    return 0;
}