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>
#include <vector>

using namespace std;

const int MOD = 1000000007;

long long modPow(long long a, long long p){
    if(p == 0) return 1;
    auto res = modPow(a, p/2);
    res = (res*res)%MOD;
    if(p%2) res = (res*a)%MOD;
    return res;
}

long long calcInv(long long a){
    return modPow(a, MOD-2);
}

int main(){
	vector<long long> inv(200002);
	vector<long long> fact(200002);
	vector<long long> factInv(200002);
	for(int i=0;i<2;i++) inv[i] = fact[i] = factInv[i] = 1;
	for(int i=2;i<=inv.size()-1;i++){
		inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
		fact[i] = fact[i-1] * i % MOD;
		factInv[i] = factInv[i-1] * inv[i] % MOD;
	}
	auto comb = [&](int n, int r){
		if(n < r) return 0LL;
		return fact[n] * factInv[n-r] % MOD * factInv[r] % MOD;
	};
    
    int N, A, B, C;
    while(cin >> N >> A >> B >> C){
        auto probA = A * inv[100 - C] % MOD;
        auto probB = B * inv[100 - C] % MOD;
        auto mul = 100 * inv[100 - C] % MOD;
        long long res = 0;
        for(int i=0;i<N;i++){
            auto pattern = comb(N-1+i, N-1);
            auto pa = modPow(probA, N-1) * modPow(probB, i) % MOD * probA % MOD;
            auto pb = modPow(probB, N-1) * modPow(probA, i) % MOD * probB % MOD;
            res += mul * (N+i) % MOD * pattern % MOD * (pa + pb) % MOD;
            res %= MOD;
        }
        cout << res << endl;
    }
}