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 <iostream>

using namespace std;
typedef long long ll;
const int MAX = 200010;
ll inv[MAX],fact[MAX],fi[MAX],mod = 1000000007;
void inverse(){
	int i;
	inv[1] = 1;
	for(i=2;i<MAX;i++){
		inv[i] = mod - (mod/i)*inv[mod%i]%mod;
	}
	fact[0] = fi[0] = 1;
	for(i=1;i<MAX;i++){
		fact[i] = fact[i-1]*i%mod;
		fi[i] = (fi[i-1]*inv[i])%mod;
	}
}

ll comb(int n,int k){
	if(n<0 || k<0 || n<k){
		return 0;
	}else{
		return fact[n]%mod*fi[k]%mod*fi[n-k]%mod;
	}
}

ll pw(ll a, ll x){
	ll ret = 1;
	a %= mod;
	while(x){
		while(!(x&1)){
			(a *= a) %= mod;
			x /= 2;
		}
		(ret *= a) %= mod;
		x--;
	}
	return ret;
}
int main(){
	ll n,a,b,c,i,j;
	cin >> n >> a >> b >> c;
	inverse();
	ll s = 0,t = 0,p = 1,q = 1;
	ll x = b*inv[100 - c]%mod,y = a*inv[100 - c]%mod;
	for(i=0;i<n;i++){
		(s += comb(n + i - 1,i)%mod*p%mod*(n + i)) %= mod;
		(p *= x) %= mod;
	}
	(s *= pw(a*inv[100-c],n)) %= mod;
	for(i=0;i<n;i++){
		(t += comb(n + i - 1,i)%mod*q%mod*(n + i)) %= mod;
		(q *= y) %= mod;
	}
	(t *= pw(b*inv[100-c],n)) %= mod;
	cout << (s + t)*100%mod*inv[100 - c]%mod << endl;
}