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>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x)  cerr << #x << " = " << x << endl

using namespace std;

typedef long long ll;
typedef string str;

const ll MAXN = 2e5 + 10;
const ll MOD = 1e9 + 7;

ll mul(ll a, ll b){
	return (a * b) % MOD;
}
ll pw(ll b, ll P){
	if(P == 0) return 1;
	if(P & 1) return mul(b, pw(b, P ^ 1));
	return pw(mul(b, b), P >> 1);
}
ll f[MAXN], iv[MAXN];
ll nCr(ll n, ll r){
	return mul(f[n], mul(iv[r], iv[n - r]));
}
ll inv(ll x){
	return pw(x, MOD - 2);
}

ll ans = 0;

void calc(ll a, ll b, ll c, ll p1, ll p2){
	ll C = mul(100, inv(100 - c));
	//debug(C);
	ll prob = mul(nCr(p1 + p2 - 1, p1 - 1), mul(   pw( mul(b, inv(a + b)), p2 ), pw( mul(a, inv(a + b)), p1 )));
	ll wait = mul(C, p1 + p2);
	ans = (ans + mul(prob, wait)) % MOD;
}

int main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	f[0] = 1;
	for(int i = 1; i < MAXN; i++) f[i] = mul(f[i - 1], i);
	iv[MAXN - 1] = pw(f[MAXN - 1], MOD - 2);
	for(int i = MAXN - 2; i >= 0; i--) iv[i] = mul(iv[i + 1], i + 1);
	
	ll n, a, b, c;
	cin >> n >> a >> b >> c;
	for(int i = 0; i < n; i++) calc(a,b,c,n,i);
	for(int i = 0; i < n; i++) calc(b,a,c,n,i);
	cout << ans;
	return 0;
}
