Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total.

When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %. Find the expected number of games that will be played, and print it as follows.

We can represent the expected value as P/Q with coprime integers P and Q. Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}. (Such an integer R always uniquely exists under the constraints of this problem.)

Constraints

* 1 \leq N \leq 100000
* 0 \leq A,B,C \leq 100
* 1 \leq A+B
* A+B+C=100
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N A B C


Output

Print the expected number of games that will be played, in the manner specified in the statement.

Examples

Input

1 25 25 50


Output

2


Input

4 50 50 0


Output

312500008


Input

1 100 0 0


Output

1


Input

100000 31 41 28


Output

104136146
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD 1000000007
#define N 200005
ll fact[N], inv[N], finv[N];
void make()
{
	fact[0] = 1;
	fact[1] = 1;
	inv[1] = 1;
	finv[1] = finv[0] = 1;
	for(int i = 2; i < N; ++i)
	{
		fact[i] = fact[i - 1] * i % MOD;
		inv[i] = MOD - (MOD / i) * inv[MOD % i] % MOD;
		finv[i] = finv[i -1] * inv[i] % MOD;
	}
}

ll C(int a, int b)
{
	if(a < b) return 0;
	return fact[a] * (finv[b] * finv[a - b] % MOD) % MOD;
}

ll exp(ll a, ll b)
{
	ll res = 1;
	while(b)
	{
		if(b & 1)
			res = res * a % MOD;
		b >>= 1;
		a = a * a % MOD;
	}
	return res;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	make();
	ll n, a, b, c;
	cin >> n >> a >> b >> c;

	ll ans = 0, pa = a * inv[a + b] % MOD, pb = b * inv[a + b] % MOD;
	for (int i = 0; i < n; ++i)
	{
		ans += C(n + i - 1, i) * exp(pa, i) % MOD * exp(pb, n) % MOD * (i + n) % MOD;
		ans += C(n + i - 1, i) * exp(pb, i) % MOD * exp(pa, n) % MOD * (i + n) % MOD;
		ans %= MOD; 
	}
	ans = ans * inv[100 -c] % MOD *100 % MOD;
	cout << ans; 
	return 0;
}