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 dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
int n, m, anses[8010], a[8010][8010], R, pi[10], ptr;
map<pair<int, int>, int> st;
int qpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % 1000000007)
    if (b & 1) ans = 1ll * ans * a % 1000000007;
  return ans;
}
int main() {
  scanf("%d", &R);
  int sum = 0;
  for (int i = 0; i < 4; ++i)
    scanf("%d", &pi[i]), sum = (sum + pi[i]) % 1000000007;
  sum = qpow(sum, 1000000007 - 2);
  for (int i = 0; i < 4; ++i) pi[i] = (long long)pi[i] * sum % 1000000007;
  for (int i = -R; i <= R; ++i)
    for (int j = -R; j <= R; ++j)
      if (i * i + j * j <= R * R) st[pair<int, int>(i, j)] = ++ptr;
  for (auto p : st) {
    a[p.second][p.second]++;
    for (int i = 0; i < 4; ++i) {
      int px = p.first.first + dx[i];
      int py = p.first.second + dy[i];
      if (st.find(pair<int, int>(px, py)) == st.end()) continue;
      a[p.second][st[pair<int, int>(px, py)]] = 1000000007 - pi[i];
    }
    a[p.second][ptr + 1]++;
  }
  for (int i = 1, j = 1; i <= ptr; ++i) {
    for (j = i; j <= ptr && !a[j][i]; ++j)
      ;
    if (j == ptr + 1) continue;
    for (int k = i; k <= ptr + 1; ++k) swap(a[j][k], a[i][k]);
    int iv = qpow(a[i][i], 1000000007 - 2);
    vector<int> f;
    for (int k = i; k <= ptr + 1; ++k)
      if (a[i][k]) f.push_back(k);
    for (int k = i; k <= ptr; ++k)
      if (i != k && a[k][i]) {
        int t = 1ll * a[k][i] * iv % 1000000007;
        for (auto p : f)
          a[k][p] =
              (a[k][p] - (long long)t * a[i][p] % 1000000007 + 1000000007) %
              1000000007;
      }
  }
  int z = st[pair<int, int>(0, 0)];
  for (int i = ptr; i >= z; --i) {
    int b = a[i][ptr + 1];
    for (int j = ptr; j >= i; --j)
      if (a[i][j])
        b = (b - (long long)a[i][j] * anses[j] % 1000000007 + 1000000007) %
            1000000007;
    anses[i] = (long long)b * qpow(a[i][i], 1000000007 - 2) % 1000000007;
  }
  printf("%d", anses[z]);
}