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>
#include <cstdio>

#define MOD 1000000007
#define SIZE 200005

using namespace std;
typedef long long ll;

ll inv[SIZE], fac[SIZE], finv[SIZE];

void com_init() {
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[1] = 1;

		for (int i = 2; i < SIZE; i++) {
				inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
				fac[i] = fac[i-1] * (ll)i % MOD;
				finv[i] = finv[i-1] * inv[i] % MOD;
		}
}

ll com(int a, int b) {
		if (a < b) return 0;
		return fac[a] * (finv[b] * finv[a-b] % MOD) % MOD;
}

ll mpow(ll x, ll t) {
		if (t == 0) return 1;
		ll ret = mpow(x, t/2);
		ret = ret * ret % MOD;
		if(t%2==1) ret = ret * x % MOD;
		
		return ret;
}

int main() {
		com_init();
		int n, A, B, DC;
		scanf("%d %d %d %d", &n, &A, &B, &DC);
		ll pa = (ll)A * inv[A+B] % MOD;
		ll pb = (ll)B * inv[A+B] % MOD;
		ll ret = 0;

		for (int k = 0; k < n; k++) {
				ret += com(k+n-1, k) * mpow(pb, k) % MOD * mpow(pa, n) % MOD * (ll)(k+n) % MOD;
				ret += com(k+n-1, k) * mpow(pa, k) % MOD * mpow(pb, n) % MOD * (ll)(k+n) % MOD;
				ret %= MOD;
		}

		ret = ret * inv[A+B] % MOD * 100LL % MOD;
		printf("%lld\n", ret);

		return 0;
}
