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<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(long long i=0;i<(long long)(n);i++)
#define N 200010


int main() {
	long long p[N];
	long long q[N];
	long long r[N];
	long long k;
	bool s[40];
	long long ss[40];
	long long n, a, b, c,d;
	long long x, y, z, ans;
	x = MOD - 2;
	f(i, 40) {
		if (x % 2 == 1)s[i] = true;
		else s[i] = false;
		x /= 2;
	}
	f(i, N) {
		ss[0] = i;
		f(j, 39)ss[j + 1] = (ss[j] * ss[j]) % MOD;
		r[i] = 1;
		f(j, 40)if (s[j])r[i] = (r[i] * ss[j]) % MOD;
	}
	scanf("%lld %lld %lld %lld", &n, &a, &b, &c);
	d = a + b;
	p[0] = 1;
	q[0] = 1;
	f(i, N-1) {
		p[i + 1] = (p[i] * a) % MOD;
		p[i + 1] = (p[i + 1] * r[d]) % MOD;
		q[i + 1] = (q[i] * b) % MOD;
		q[i + 1] = (q[i + 1] * r[d]) % MOD;
	}
	ans = 0;
	x = n;
	f(i, n) {
		y = (p[n - i] + q[n - i]) % MOD;
		y = (y*p[i]) % MOD;
		y = (y*q[i]) % MOD;
		y = (y*x) % MOD;
		ans = (ans + y) % MOD;
		if (i < (n - 1)) {
			x = (x*(n + i + 1)) % MOD;
			x = (x*r[i + 1]) % MOD;
		}
}
	ans = (ans * 100) % MOD;
	ans = (ans*r[d]) % MOD;
	printf("%lld\n", ans);


	return 0;
}