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;
int n, A, B, w[105][105], cnt, f[8000], s[8000][205], val[8000], ans[8000],
    p[5], sum;
int tx[] = {-1, 0, 1, 0}, ty[] = {0, -1, 0, 1};
int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * x % 1000000007;
    x = (long long)x * x % 1000000007;
    y >>= 1;
  }
  return ret;
}
int gauss() {
  int i, j, k;
  for (i = 1; i <= cnt; i++) {
    for (j = i + 1; j <= cnt && (j - i) <= A; j++)
      if (s[j][A - (j - i)]) {
        int num = (long long)s[j][A - (j - i)] * qpow(s[i][A], 1000000007 - 2) %
                  1000000007;
        for (k = A; k <= B; k++) {
          s[j][k - (j - i)] =
              (s[j][k - (j - i)] - (long long)num * s[i][k] % 1000000007 +
               1000000007) %
              1000000007;
        }
        val[j] = (val[j] - (long long)num * val[i] % 1000000007 + 1000000007) %
                 1000000007;
      }
  }
  for (i = cnt; i >= w[n][n]; i--) {
    ans[i] = (long long)val[i] * qpow(s[i][A], 1000000007 - 2) % 1000000007;
    for (int j = i - 1; j && i - j <= A; j--)
      if (s[j][A + (i - j)])
        val[j] = (val[j] - (long long)ans[i] * s[j][A + (i - j)] % 1000000007 +
                  1000000007) %
                 1000000007;
  }
  return ans[w[n][n]];
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  A = n * 2;
  B = n * 4;
  for (i = 0; i < 4; i++) {
    scanf("%d", p + i);
    sum += p[i];
  }
  for (i = 0; i <= A; i++)
    for (j = 0; j <= A; j++)
      if ((i - n) * (i - n) + (j - n) * (j - n) <= n * n) w[i][j] = ++cnt;
  for (i = 0; i <= A; i++)
    for (j = 0; j <= A; j++)
      if (w[i][j]) {
        for (k = 0; k < 4; k++) {
          int x = i + tx[k], y = j + ty[k];
          if (w[x][y]) s[w[i][j]][w[x][y] - w[i][j] + A] = p[k];
        }
        s[w[i][j]][A] = 1000000007 - sum;
        val[w[i][j]] = 1000000007 - sum;
      }
  printf("%d\n", gauss());
  return 0;
}