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 <stdio.h>
#include <algorithm>

using namespace std;

//long long
const long long mod = 1000000007;
long long n , a , b , c;
long long pwa[120000] , pwb[120000];
long long fac[420000] , ifac[420000];
long long ans;
long long ksm ( long long f , long long x ) {
	long long s = 1;
	while ( x ) {
		if ( x % 2 ) s = s * f % mod;
		f = f * f % mod; x = x / 2;
	}
	return s % mod;
}
long long inv ( long long x ) {
	return ksm ( x , mod - 2 );
}
long long C ( long long x , long long y ) {
	return fac[x] * ifac[y] % mod * ifac[x-y] % mod;
}
void work () {
	long long i;
	scanf ( "%lld%lld%lld%lld" , &n , &a , &b , &c );
	a = a * inv ( 100 - c ) % mod;
	b = b * inv ( 100 - c ) % mod;
	c = 100 * inv ( 100 - c ) % mod;
	fac[0] = 1;
	for ( i = 1 ; i <= 400000 ; i++ ) {
		fac[i] = fac[i-1] * i % mod;
	}
	ifac[400000] = inv ( fac[400000] );
	for ( i = 399999 ; i >= 0 ; i-- ) {
		ifac[i] = (ifac[i+1]*(i+1)) % mod;
	}
	pwa[0] = pwb[0] = 1;
	for ( i = 1 ; i <= n ; i++ ) {
		pwa[i] = pwa[i-1] * a % mod;
		pwb[i] = pwb[i-1] * b % mod;
	}
	for ( i = 0 ; i <= n - 1 ; i++ ) {
		ans = (ans + C(n+i-1,n-1)*((pwa[n]*pwb[i]+pwb[n]*pwa[i]) % mod)%mod * (n+i) % mod) % mod;
	}
	if ( c == 0 ) c = 1;
	printf ( "%lld\n" , ans * c % mod );
}
int main () {
	work ();
	return 0;
}
