#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
using namespace std;
typedef long long ll;
const int N = 30000010;
const ll mod = 1000000007, inv2 = (mod+1)>>1;
ll qpow(ll x, ll y = mod - 2){
	ll ret = 1;
	for(; y; y >>= 1, (x *= x) %= mod) if(y & 1) (ret *= x) %= mod;
	return ret;
}
ll fac[N], ifac[N];

ll C(ll x, ll y){
	if(x < y || y < 0) return 0;
	return fac[x] * ifac[x-y] % mod * ifac[y] % mod;
}

int n;
ll m, ans;
void solve(){
	scanf("%d%lld", &n, &m);
	if(!m) {puts("1"); return;}
	if(n < m) {puts("0"); return;}
	if(n == m) {printf("%lld\n", fac[n-1] * inv2 % mod); return;}
	ll pw = 1;
	ans = 0;
	rep(i, 1, n-m){
		(pw *= inv2) %= mod;
		(ans += C(n-m, i) * C(m-1, i-1) % mod * pw) %= mod;
	}
	(ans *= fac[n] * ifac[n-m] % mod) %= mod;
	printf("%lld\n", ans);
}

int main(){
	fac[0] = 1; rep(i, 1, N-5) fac[i] = fac[i-1] * i % mod;
	ifac[N-5] = qpow(fac[N-5]);
	per(i, N-6, 0) ifac[i] = ifac[i+1] * (i+1) % mod;
	int T; scanf("%d", &T); while(T--) solve();
	return 0;
}