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>
#define llint long long
#define mod 1000000007

using namespace std;

const int FACT_MAX = 200005;
llint fact[FACT_MAX], fact_inv[FACT_MAX];

llint modpow(llint a, llint n)
{
	if(n == 0) return 1;
	if(n % 2){
		return ((a%mod) * (modpow(a, n-1)%mod)) % mod;
	}
	else{
		return modpow((a*a)%mod, n/2) % mod;
	}
}

void make_fact()
{
	llint val = 1;
	fact[0] = 1;
	for(int i = 1; i < FACT_MAX; i++){
		val *= i;
		val %= mod;
		fact[i] = val;
	}
	fact_inv[FACT_MAX-1] = modpow(fact[FACT_MAX-1], mod-2);
	for(int i = FACT_MAX-2; i >= 0; i--){
		fact_inv[i] = fact_inv[i+1] * (i+1) % mod;
	}
}

llint comb(llint n, llint k)
{
	llint ret = 1;
	ret *= fact[n];
	ret *= fact_inv[k], ret %= mod;
	ret *= fact_inv[n-k], ret %= mod;
	return ret;
}

llint n, a, b, c;

int main(void)
{
  cin >> n >> a >> b >> c;
  make_fact();
  a *= modpow(100, mod-2), a %= mod;
  b *= modpow(100, mod-2), b %= mod;
  c *= modpow(100, mod-2), c %= mod;

  llint ans = 0;
  llint A = a * modpow((a+b)%mod, mod-2) % mod, B = b * modpow((a+b)%mod, mod-2) % mod;
  for(int i = n; i <= 2*n-1; i++){
    llint tmp = comb(i-1, n-1) * modpow(A, n) % mod * modpow(B, i-n) % mod;
    tmp += comb(i-1, n-1) * modpow(B, n) % mod * modpow(A, i-n) % mod, tmp %= mod;
    tmp *= i, tmp %= mod;
    ans += tmp, ans %= mod;
  }
  ans *= modpow((a+b)%mod, mod-2), ans %= mod;
  cout << ans << endl;
  return 0;
}
