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 main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	const int mod=1e9+7;
	auto add = [&mod](int a, int b) -> int {a += b; return a >= mod? a - mod: a;};
	auto sub = [&mod](int a, int b) -> int {a -= b; return a < 0? a + mod: a;};
	auto mul = [&mod](int a, int b) -> int {return 1LL * a * b % mod;};
	auto qpow = [&](int64_t t,int64_t p) -> int {
		int64_t ans = 1;
		while (p) {
			if(p & 1) ans = ans * t %mod;
			t = t * t % mod;
			p >>=1;
		}
		return ans;
 	};
	int n, a, b, c;
	cin >> n >> a >> b >> c;
	vector<int> pre(2 * n + 1), inv(2 * n + 1);
	pre[0] = 1;
	inv[0] = 1;
	for(int i = 1; i <= 2 * n; ++i) {
		pre[i] = mul(pre[i - 1], i);
		inv[i] = qpow(pre[i], mod-2);
	}
	auto C = [&](int n, int m) {
		return mul(pre[n], mul(inv[m], inv[n-m]));
	};
	int ans = 0;
	int inv100 = qpow(100, mod - 2);
	a = mul(a, inv100);
	b = mul(b, inv100);
	c = mul(c, inv100);
	int tmp = qpow(sub(1, c), mod - 2);
	for (int i = 0; i < n; ++i) {
		int tt = mul(C(i + n - 1, n - 1), qpow(b, i));
		tt = mul(tt, qpow(a, n));
		tt = mul(tt, n + i);
		ans = add(ans, mul(tt, qpow(tmp, n + i + 1)));
	}
	for (int i = 0; i < n; ++i) {
		int tt = mul(C(i + n - 1, n - 1), qpow(a, i));
		tt = mul(tt, qpow(b, n));
		tt = mul(tt, n + i);
		ans = add(ans, mul(tt, qpow(tmp, n + i + 1)));
	}
	cout << ans <<endl;
	return 0;
}