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 maxn = 8000, p = 1e9 + 7;
int sz = 0, r, cnt;
int mem[110][110];
int a[8500][110];
int b[110][110];
int get(int i, int j) {
  if (mem[i + 50][j + 50]) return mem[i + 50][j + 50];
  if (i * i + j * j > r * r) return 0;
  return mem[i + 50][j + 50] = ++sz;
}
int power_mod(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) {
      res = 1ll * res * a % p;
    }
    a = 1ll * a * a % p;
    n >>= 1;
  }
  return res;
}
void gauss(int a[][110], int sz) {
  for (int i = 1; i <= sz; i++) {
    int inv = power_mod(a[i][i], p - 2) % p;
    for (int k = i + 1; k <= sz; k++) {
      int now = 1ll * a[k][i] * inv % p;
      for (int j = sz + 1; j >= i; j--) {
        a[k][j] = (a[k][j] - 1ll * a[i][j] * now % p + p) % p;
      }
    }
  }
  for (int i = sz; i >= 1; i--) {
    int x = a[i][sz + 1] = 1ll * a[i][sz + 1] * power_mod(a[i][i], p - 2) % p;
    for (int k = 1; k < i; k++) {
      a[k][sz + 1] = (a[k][sz + 1] - 1ll * x * a[k][i] % p + p) % p;
    }
  }
}
int nx;
int a1, a2, a3, a4;
void getfun(int i, int j, int ty) {
  int num;
  if (ty == 1) {
    num = get(i + 1, j);
  } else {
    num = ++cnt;
  }
  int t = get(i, j), t1 = get(i - 1, j), t2 = get(i, j - 1), t4 = get(i, j + 1);
  for (int k = 0; k <= nx; k++) {
    a[num][k] = (1ll * a[t][k] + 1ll * a[t1][k] * a1 % p +
                 1ll * a[t2][k] * a2 % p + 1ll * a[t4][k] * a4 % p) %
                p;
  }
  a[num][0] -= 1;
  if (a[num][0] < 0) a[num][0] += p;
  if (!ty) return;
  for (int k = 0; k <= nx; k++) {
    a[num][k] = 1ll * a[num][k] * a3 % p;
  }
}
int main() {
  scanf("%d%d%d%d%d", &r, &a1, &a2, &a3, &a4);
  int t = power_mod(a1 + a2 + a3 + a4, p - 2);
  a1 = p - 1ll * a1 * t % p, a2 = p - 1ll * a2 * t % p,
  a4 = p - 1ll * a4 * t % p;
  a3 = power_mod(1ll * a3 * t % p, p - 2);
  for (int j = r; j >= -r; j--) {
    for (int i = -r; i <= r; i++) {
      if (get(i, j)) {
        a[get(i, j)][get(i, j)] = 1;
        break;
      }
    }
  }
  nx = 2 * r + 1;
  for (int i = -r; i <= r; i++) {
    for (int j = r; j >= -r; j--) {
      if (!mem[i + 50][j + 50] && get(i, j)) {
        getfun(i - 1, j, 1);
      }
    }
  }
  cnt = sz;
  for (int j = r; j >= -r; j--) {
    for (int i = 0; i <= r; i++) {
      if (get(i, j) && !get(i + 1, j)) {
        getfun(i, j, 0);
        break;
      }
    }
  }
  for (int i = sz + 1; i <= cnt; i++) {
    for (int j = 1; j <= nx; j++) {
      b[i - sz][j] = a[i][j];
    }
    b[i - sz][nx + 1] = p - a[i][0];
  }
  gauss(b, nx);
  int num = get(0, 0);
  int ans = a[num][0];
  for (int i = 1; i <= nx; i++) {
    ans += 1ll * a[num][i] * b[i][nx + 1] % p;
    ans -= ans >= p ? p : 0;
  }
  printf("%d\n", ans);
  return 0;
}