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#include 
   
    typedef long long ll;const int N = 100005;		// 最大能计算的n上限const int MOD = 1000000007; // 1e9 + 7ll fac[N]; // 用于存储阶乘结果ll qpow(ll a, ll b) { // 快速幂	ll res = 1, base = a;	while (b) {		if (b & 1) res = res * base % MOD;		base = base * base % MOD;		b >>= 1;	}	return res;}void init() {	fac[0] = 1;	for (int i = 1; i < N; i++) {		fac[i] = fac[i - 1] * i % MOD;	}}ll cal(ll n, ll m) {    if (n < m) return 0;    return 1LL * fac[n] * qpow(fac[m], MOD - 2) % MOD * qpow(fac[n - m], MOD - 2) % MOD;}int main() {	init();	int n;	scanf("%d", &n);	for (int i = 0; i < n; i++) {		ll n, m;		scanf("%lld %lld", &n, &m);		ll res = cal(n, m);		printf("%lld\n", res);	}	return 0;}