下一盘赢的概率p, 输的概率1-p, 期望赢p-(1-p)=2p-1盘. 所以赢2盘的期望为$\frac{2}{2p-1}$.

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;

i64 qmi(i64 a, i64 m, i64 p)
{
	i64 ans = 1;
	while (m)
	{
		if (m & 1) (ans *= a) %= p;
		(a *= a) %= p;
		m >>= 1;
	}
	return ans;
}

i64 inv(i64 x)
{
	static const i64 mod = 1e9 + 7;
	return qmi(x % mod, mod - 2, mod);
}

void solve()
{
	int n, m; cin >> n >> m;
	static const i64 mod = 1e9 + 7;	
	i64 ans = (n * 2ll) % mod;
	(ans *= inv(m * 2ll - n)) %= mod;
	cout << ans << "\n";
}

int main()
{
	int T = 1; cin >> T;
	while (T--) solve();	
	return 0;
}