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;

const int maxn = 1e5+10;
const long long M = 1e9+7;
typedef long long ll;

int n, a, b, c, powa[maxn], powb[maxn], inv[maxn], Cn[maxn<<1];
long long ans;

int main() {
	scanf("%d%d%d%d", &n, &a, &b, &c);
	inv[1] = 1;
	for (int i = 2; i < maxn; ++i) {
		inv[i] = (M-(M/i)*inv[M%i]%M)%M;
	}
	powa[0] = powb[0] = 1;
	for (int i = 1; i <= n; ++i) {
		powa[i] = (ll)powa[i-1]*a%M*inv[a+b]%M;
		powb[i] = (ll)powb[i-1]*b%M*inv[a+b]%M;
	}
	Cn[n-1] = 1;
	for (int i = n; i < 2*n; ++i) {
		Cn[i] = (ll)Cn[i-1]*i%M*inv[i-n+1]%M;
	}
	for (int i = n; i < 2*n; ++i) {
		ans = (ans+(ll)i*Cn[i-1]%M*(((ll)powa[n]*powb[i-n]+(ll)powb[n]*powa[i-n])%M))%M;
	}
	ans = ans*100%M*inv[100-c]%M;
	printf("%lld\n", ans);
}
