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;

#define dump(x)  cout << #x << " = " << (x) << endl;

typedef long long ll;
typedef pair<int, int> ii;
const int mod = 1e9+7;
const ll INF = 1e18;
const int inf = 1e9;

const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];

void cominit()
{
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i=2; i<MAX; i++) {
    fac[i] = fac[i-1] * i % mod;
    inv[i] = mod - inv[mod%i] * (mod/i) % mod;
    finv[i] = finv[i-1] * inv[i] % mod;
  }
}

ll com(ll n, ll k)
{
  if (n<k) return 0;
  if (n<0 || k<0) return 0;
  return fac[n] * (finv[k] * finv[n-k] % mod) % mod;
}

ll modpow(ll a, ll n)
{
  ll res=1;
  while (n>0) {
    if (n&1) res = res*a % mod;
    a = a*a % mod;
    n >>= 1;
  }
  return res;
}

void solve()
{
  cominit();
  ll n,a,b,c; cin>>n>>a>>b>>c;

  ll pa = a * inv[a+b] % mod;
  ll pb = b * inv[a+b] % mod;

  ll ans = 0;
  for (int i=0; i<n; i++) {
    ans += modpow(pa,n) * modpow(pb,i) % mod * com(n-1+i, i) % mod * (i+n) % mod;
    ans += modpow(pb,n) * modpow(pa,i) % mod * com(n-1+i, i) % mod * (i+n) % mod;
    ans %= mod;
  }
  ans = ans * inv[a+b] % mod * 100LL % mod;
  cout << ans << endl;
}

int main()
{
  solve();
  return 0;
}
