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;

#define rep(i, n) for(int i = 0; i < (int)n; i++)

typedef long long ll;

ll mod = 1000000007;

ll kai[200010];
ll inv[200010];
ll aa[100010], bb[100010];

ll beki(ll x, int y)
{
	if(y==0) return 1;
	if(y==1) return x;
	ll tmp=beki(x, y/2);
	if(y%2==0) return tmp*tmp%mod;
	else return tmp*tmp%mod*x%mod;
}

ll inverse(ll x)
{
	return beki(x, mod-2);
}

int main()
{
	ll n, a, b, c;
	cin >> n >> a >> b >> c;
	ll r = 100*inverse(100-c)%mod;
	ll tmp = inverse(a+b);
	b=b*tmp%mod;
	a=a*tmp%mod;
	aa[0]=1;
	bb[0]=1;
	rep(i, n+1)
	{
		aa[i+1]=aa[i]*a%mod;
		bb[i+1]=bb[i]*b%mod;
	}
	kai[0]=1;
	inv[0]=1;
	rep(i, 2*n+1)
	{
		kai[i+1]=kai[i]*((ll)i+1)%mod;
		inv[i+1]=inverse(kai[i+1]);
	}
	ll ans=0;
	rep(k, n)
	{
		ans+=kai[n+k]*inv[n-1]%mod*inv[k]%mod*aa[n]%mod*bb[k]%mod;
		ans%=mod;
		ans+=kai[n+k]*inv[n-1]%mod*inv[k]%mod*bb[n]%mod*aa[k]%mod;
		ans%=mod;
	}
	ans=ans*r%mod;
	cout << ans << endl;
	return 0;
}