#include <bits/stdc++.h>
using namespace std;
const int maxn      = 2e6+233;
typedef long long LL;
const LL mod = 1e9+7;
LL T,n,m,c,ok,tmp,add,ret,ans,sum,tot;
LL arr[maxn],fac[maxn],fac2[maxn];
LL qpow(LL a, LL n)//计算a^n % mod
{
    LL ans  = 1;
    while(n)
    {
        if(n & 1)//判断n的最后一位是否为1
            ans = (ans * a) % mod;
        n >>= 1;//舍去n的最后一位
        a = (a * a) % mod;//将a平方
    }
    return ans % mod;
}
LL comF(LL a,LL b)
{
    return ((fac[a]*fac2[b]%mod)*(fac2[a-b])%mod);
}
void init()
{
    fac[0]=1;
    for(LL i=1; i<=200000; ++i)
    {
        fac[i]=(fac[i-1]*i)%mod;
    }
    fac2[200000]=qpow(fac[200000],mod-2);
    for(int i=199999; i>=0; --i)
        fac2[i]=(fac2[i+1]*(i+1))%mod;
}
LL cal(LL x,LL y)
{
    return comF(x,y)%mod;
}
LL solve()
{
    ans=1;
    tot=1;
    tmp=0;
    LL i;
    for(i=2; i<=n*2; ++i)
    {
        if(arr[i]!=arr[i-1])
        {
            if(tmp==1)
            {
                if(i%2==0)
                {
                    ans=(ans*(cal(tot,tot/2)+cal(tot,tot/2+1)))%mod;
                }
                else
                {
                    ans=(ans*cal(tot,tot/2))%mod;
                }
            }
            else
            {
                if(tot%2)
                    ans=(ans*(cal(tot,tot/2)+cal(tot,tot-tot/2)))%mod;
                else
                    ans=(ans*cal(tot,tot/2))%mod;
            }
            if(i%2==0)
                tmp=1;
            else
                tmp=0;
            tot=1;
        }
        else
            tot++;
    }
    if(tmp==1)
    {
        if(i%2==0)
        {
            ans=(ans*(cal(tot,tot/2)+cal(tot,tot/2+1)))%mod;
        }
        else
        {
            ans=(ans*cal(tot,tot/2))%mod;
        }
    }
    else
    {
        if(tot%2)
            ans=(ans*(cal(tot,tot/2)+cal(tot,tot-tot/2)))%mod;
        else
            ans=(ans*cal(tot,tot/2))%mod;
    }
    return ans;
}
int main()
{
   // freopen("in.txt","r",stdin);
    init();
    scanf("%I64d",&T);
    while(T--)
    {
        scanf("%I64d",&n);
        for(LL i=1; i<=n*2; ++i)
        {
            scanf("%I64d",&arr[i]);
        }
        ret = solve();
        printf("%I64d\n",ret);
    }
    return 0;
}