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 Mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
int n, t, sz, p[4], f[8005][8005], h[110000];
inline int hs(int x, int y) {
  x += (n + 1);
  y += (n + 1);
  return (x - 1) * t + y;
}
inline int dis2(int x, int y) { return x * x + y * y; }
inline void MINUS(int &x, int y) {
  x -= y;
  if (x < 0) x += Mod;
}
long long qpow(long long x, int a) {
  long long res = 1;
  while (a) {
    if (a & 1) res = res * x % Mod;
    x = x * x % Mod;
    a >>= 1;
  }
  return res;
}
void Gauss() {
  for (int i = 1; i <= sz; i++) {
    int inv = qpow(f[i][i], Mod - 2);
    for (int j = i + 1; j <= min(i + 1 + t, sz); j++)
      f[i][j] = 1ll * f[i][j] * inv % Mod;
    f[i][sz + 1] = 1ll * f[i][sz + 1] * inv % Mod;
    f[i][i] = 1ll;
    for (int j = i + 1; j <= min(i + 1 + t, sz); j++) {
      if (!f[j][i]) continue;
      for (int k = i + 1; k <= min(i + 1 + t, sz); k++)
        MINUS(f[j][k], 1ll * f[j][i] * f[i][k] % Mod);
      MINUS(f[j][sz + 1], 1ll * f[j][i] * f[i][sz + 1] % Mod);
      f[j][i] = 0ll;
    }
  }
  int to = h[hs(0, 0)];
  for (int i = sz; i >= to; i--) {
    int inv = qpow(f[i][i], Mod - 2);
    f[i][sz + 1] = 1ll * f[i][sz + 1] * inv % Mod;
    for (int j = i - 1; j >= to; j--)
      MINUS(f[j][sz + 1], 1ll * f[j][i] * f[i][sz + 1] % Mod);
  }
}
int main() {
  scanf("%d%d%d%d%d", &n, &p[0], &p[1], &p[2], &p[3]);
  t = 2 * n + 1;
  int lim = n * n, s = p[0] + p[1] + p[2] + p[3];
  sz = 0;
  for (int i = -n; i <= n; i++)
    for (int j = -n; j <= n; j++)
      if (dis2(i, j) <= lim) h[hs(i, j)] = ++sz;
  for (int i = -n; i <= n; i++)
    for (int j = -n; j <= n; j++) {
      int now = h[hs(i, j)];
      if (dis2(i, j) <= lim) {
        MINUS(f[now][now], s);
        MINUS(f[now][sz + 1], s);
        for (int k = 0; k < 4; k++) {
          int x = i + dx[k], y = j + dy[k];
          if (dis2(x, y) <= lim) f[now][h[hs(x, y)]] = p[k];
        }
      }
    }
  Gauss();
  printf("%d\n", f[h[hs(0, 0)]][sz + 1]);
  return 0;
}