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;

int n, a, b, c;
const int mod = 1e9 + 7;
typedef long long ll;
ll expmod(ll a, ll b, ll mod) {
	ll rlt = 1;
	for (a %= mod; b; a = a * a % mod, b >>= 1) if (b & 1) rlt = rlt * a % mod;
	return rlt;
}

#define N 200020
int fac[N], rfac[N], ri[N];
void init() {
	fac[0] = 1;
	for (int i = 1; i < N; i ++) fac[i] = 1ll * fac[i-1] * i % mod;
	rfac[0] = rfac[1] = 1, ri[1] = 1;
   	for(int i = 2; i < N; i ++){
      		int t = mod / i;
      		ri[i] = (1ll * t * (mod - ri[mod % i])) % mod;
      		rfac[i] = (1ll * rfac[i-1] * ri[i]) % mod;
   	}
}

int func(int a, int b, int c) {
	int r = expmod(a + b, mod - 2, mod);
	int now = expmod(r, n + 1, mod);
	int sum = 0;
	for (int i = 0; i < n; i ++) {
		sum = (sum + 1ll * fac[n+i] * rfac[i] % mod * now) % mod;
		now = 1ll * now * b % mod * r % mod;
	}
	return 1ll * sum * expmod(a, n, mod) % mod * rfac[n-1] % mod;
}

int main() {
	init();
	scanf("%d%d%d%d", &n, &a, &b, &c);
	int q = expmod(100, mod - 2, mod);
	a = 1ll * a * q % mod;
	b = 1ll * b * q % mod;
	c = 1ll * c * q % mod;
	printf("%d\n", (func(a, b, c) + func(b, a, c)) % mod);
	return 0;
}
