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>.
#include <bits/stdc++.h>
using namespace std;
void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(int x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const int maxn = 1e6 + 10;
const int inf = 1e9;
const double eps = 1e-8;
const int mod = 1e9 + 7;
int a, b, c, d, r, f[7900][7900], id[120][120], cnt, w;
int qpow(int a, int x) {
  int res = 1;
  for (; x; x >>= 1, a = 1ll * a * a % mod)
    if (x & 1) res = 1ll * res * a % mod;
  return res;
}
void solve() {
  for (int i = 1; i <= cnt; i++)
    for (int j = i + 1; j <= min(cnt, i + r * 2); j++) {
      if (!f[j][i]) continue;
      int t = 1ll * f[j][i] * qpow(f[i][i], mod - 2) % mod;
      for (int k = i; k <= min(cnt, i + r * 2); k++)
        f[j][k] = (f[j][k] - 1ll * t * f[i][k] % mod + mod) % mod;
      f[j][cnt + 1] =
          (f[j][cnt + 1] - 1ll * t * f[i][cnt + 1] % mod + mod) % mod;
    }
  for (int i = cnt; i; i--) {
    for (int j = i + 1; j <= cnt; j++)
      f[i][cnt + 1] =
          (f[i][cnt + 1] - 1ll * f[i][j] * f[j][j] % mod + mod) % mod;
    f[i][i] = 1ll * f[i][cnt + 1] * qpow(f[i][i], mod - 2) % mod;
  }
  int x = id[w][w];
  printf("%d\n", f[x][x]);
}
int main() {
  read(r), read(a), read(b), read(c), read(d);
  int s = qpow(a + b + c + d, mod - 2);
  w = r + 2;
  a = 1ll * a * s % mod, b = 1ll * b * s % mod, c = 1ll * c * s % mod,
  d = 1ll * d * s % mod;
  for (int i = r; i >= -r; i--)
    for (int j = -r; j <= r; j++) {
      if (i * i + j * j > r * r) continue;
      id[i + w][j + w] = ++cnt;
    }
  for (int i = r; i >= -r; i--)
    for (int j = -r; j <= r; j++) {
      if (i * i + j * j > r * r) continue;
      int x = id[i + w][j + w], t;
      f[x][x] = 1;
      t = id[i - 1 + w][j + w];
      if (t) f[x][t] = mod - a;
      t = id[i + w][j + w - 1];
      if (t) f[x][t] = mod - b;
      t = id[i + 1 + w][j + w];
      if (t) f[x][t] = mod - c;
      t = id[i + w][j + w + 1];
      if (t) f[x][t] = mod - d;
      f[x][cnt + 1] = 1;
    }
  solve();
  return 0;
}
