New Year and Arbitrary Arrangement

给定三个数 $k,pa,pb$。 每次有 $\dfrac{pa}{pa+pb}$ 的概率往后面添加一个 `a`。 每次有 $\dfrac{pb}{pa+pb}$ 的概率往后面添加一个 `b`。 当出现了 $k$ 个形如 $ab$ 的子序列（不用连续）时停止。 求最后 `ab` 序列的期望数。 答案对 $10^9+7$ 取模。 Translated by yybyyb

You are given three integers $ k $ , $ p_{a} $ and $ p_{b} $ . You will construct a sequence with the following algorithm: Initially, start with the empty sequence. Each second, you do the following. With probability $ p_{a}/(p_{a}+p_{b}) $ , add 'a' to the end of the sequence. Otherwise (with probability $ p_{b}/(p_{a}+p_{b}) $ ), add 'b' to the end of the sequence. You stop once there are at least $ k $ subsequences that form 'ab'. Determine the expected number of times 'ab' is a subsequence in the resulting sequence. It can be shown that this can be represented by $ P/Q $ , where $ P $ and $ Q $ are coprime integers, and . Print the value of .

The first line will contain three integers integer $ k,p_{a},p_{b} $ ( $ 1<=k<=1000 $ , $ 1<=p_{a},p_{b}<=1000000 $ ).

Print a single integer, the answer to the problem.

1 1 1

2

3 1 4

370000006

The first sample, we will keep appending to our sequence until we get the subsequence 'ab' at least once. For instance, we get the sequence 'ab' with probability 1/4, 'bbab' with probability 1/16, and 'aab' with probability 1/8. Note, it's impossible for us to end with a sequence like 'aabab', since we would have stopped our algorithm once we had the prefix 'aab'. The expected amount of times that 'ab' will occur across all valid sequences is 2. For the second sample, the answer is equal to .