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;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int INF = 1<<30;
const ll mod = 1000000007LL;

ll pow(ll a, ll n, ll m) {
	ll ret = 1LL;
	for (; n > 0LL; n >>= 1LL, a = a * a % m) {
		if (n % 2LL == 1LL) {
			ret = ret * a % m;
		}
	}
	return ret;
}

int main() {
  ll N,A,B,C;
  cin>>N>>A>>B>>C;
  ll ans = 0LL;
  ll comb = 1LL;
  ll Q = pow(A+B,N,mod);
  for(ll i = 0LL; i < N; i++){
    ll P = (pow(A,N,mod)*pow(B,i,mod))%mod;
    P=(P*comb)%mod;
    ll R = (P*pow(Q,mod-2,mod))%mod;
    ll S = (100LL*(N+i)*pow(A+B,mod-2,mod))%mod;
    ll a = (R*S)%mod;
    P = (pow(B,N,mod)*pow(A,i,mod))%mod;
    P=(P*comb)%mod;
    R = (P*pow(Q,mod-2,mod))%mod;
    ll b = (R*S)%mod;
    ans=(ans+a+b)%mod;
    comb=(comb*(N+i))%mod;
    comb=(comb*pow(1+i,mod-2,mod))%mod;
    Q=(Q*(A+B))%mod;
  }
  cout<<ans<<endl;
}