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;

typedef long long ll;
typedef long double ld;
typedef pair<int  ,int > pii;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn  = 3e6;
const ll mod =1e9+7;
const ld PI = acos((ld)-1);

#define pb push_back
#define endl '\n'
#define dokme(x) return(cout << x , 0);
#define migmig ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ms(x , y) memset(x , y , sizeof x);
#define file_init freopen("input.txt", "r+", stdin); freopen("output.txt", "w+", stdout);
ll pw(ll a, ll b, ll md = mod){ll res = 1;while(b){if(b&1){res=(a*res)%md;}a=(a*a)%md;b>>=1;}return(res);}

ll fact[maxn] , inv[maxn];
ll n , a, b , c;

ll C(ll n ,ll r){
    return(fact[n] * inv[r] % mod * inv[n-r] % mod);
}

ll solve (){
    ll ans = 0;
    for(int i = n ; i <= 2*n - 1 ; i ++)
        ans = (ans + fact[i] * inv[n-1] %mod* inv[i-n] %mod* pw(b , i-n)%mod * pw(pw((1-c + mod)%mod , mod-2) , i + 1 ) %mod )%mod;
    ans = ans * pw(a , n) % mod;
    swap(a , b);
    return (ans);
}

int main(){
    migmig
    fact[0] = inv[0] =1;
    for(ll i = 1;  i <= 1000000 ; i ++)
        fact[i] = (i * fact[i-1])%mod , inv[i] = pw(fact[i] , mod - 2);
    cin >> n >> a >> b >> c;
    a = a * pw(100 , mod - 2) % mod;
    b = b * pw(100 , mod - 2) % mod;
    c = c * pw(100 , mod - 2) % mod;
    cout << (solve() + solve())%mod;
    return(0);
}