Problem: A chip was placed on a field with coordinate system onto point (0, 0).

Every second the chip moves randomly. If the chip is currently at a point (x, y), after a second it moves to the point (x - 1, y) with probability p1, to the point (x, y - 1) with probability p2, to the point (x + 1, y) with probability p3 and to the point (x, y + 1) with probability p4. It's guaranteed that p1 + p2 + p3 + p4 = 1. The moves are independent.

Find out the expected time after which chip will move away from origin at a distance greater than R (i.e. <image> will be satisfied).

Input

First line contains five integers R, a1, a2, a3 and a4 (0 ≤ R ≤ 50, 1 ≤ a1, a2, a3, a4 ≤ 1000).

Probabilities pi can be calculated using formula <image>.

Output

It can be shown that answer for this problem is always a rational number of form <image>, where <image>.

Print P·Q - 1 modulo 109 + 7. 

Examples

Input

0 1 1 1 1


Output

1

Input

1 1 1 1 1


Output

666666674

Input

1 1 2 1 2


Output

538461545

Note

In the first example initially the chip is located at a distance 0 from origin. In one second chip will move to distance 1 is some direction, so distance to origin will become 1.

Answers to the second and the third tests: <image> and <image>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 10010, V = 55, p = 1e9 + 7;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int R, pr[4], apr, id[V << 1][V << 1], n;
unordered_map<int, int> mp[N];
inline void A(int &x, int y) {
  if ((x += y) >= p) x -= p;
}
inline int Qmi(int x, int y = p - 2) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % p)
    if (y & 1) ans = 1ll * ans * x % p;
  return ans;
}
inline void Update(int x, int y) {
  if (!mp[y].count(x)) return;
  int k = mp[y][x];
  for (unordered_map<int, int>::const_iterator it = mp[x].begin();
       it != mp[x].end(); ++it) {
    unordered_map<int, int>::iterator ot = mp[y].find(it->first);
    if (ot == mp[y].end()) ot = mp[y].insert(make_pair(it->first, 0)).first;
    A(ot->second, p - 1ll * it->second * k % p);
    if (!ot->second) mp[y].erase(ot);
  }
}
inline void Gause() {
  int ip = id[V][V];
  for (int i = 1; i <= n; ++i) {
    if (!mp[i].count(i))
      for (int j = i + 1; j <= n; ++j)
        if (mp[j].count(i)) mp[i].swap(mp[j]);
    int ni = Qmi(mp[i][i]);
    for (unordered_map<int, int>::iterator it = mp[i].begin();
         it != mp[i].end(); ++it)
      it->second = 1ll * it->second * ni % p;
    if (i > ip) Update(i, ip);
    for (int j = i + 1; j <= n && j <= i + (R << 1); ++j) Update(i, j);
  }
  printf("%d\n", mp[ip][n + 1]);
}
int main() {
  scanf("%d", &R);
  for (int i = 0; i < 4; ++i) scanf("%d", pr + i), A(apr, pr[i]);
  apr = Qmi(apr);
  for (int i = 0; i < 4; ++i) pr[i] = 1ll * pr[i] * apr % p;
  for (int i = -R; i <= R; ++i)
    for (int j = -R; j <= R; ++j)
      if (i * i + j * j <= R * R) id[i + V][j + V] = ++n;
  for (int i = -R, no, to; i <= R; ++i)
    for (int j = -R; j <= R; ++j)
      if ((no = id[i + V][j + V])) {
        mp[no][no] = 1;
        mp[no][n + 1] = 1;
        for (int d = 0; d < 4; ++d)
          if ((to = id[i + V + dx[d]][j + V + dy[d]])) mp[no][to] = p - pr[d];
      }
  Gause();
  return 0;
}