Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."

Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?

Input

The first line contains a single integer k (1 ≤ k ≤ 109).

Output

You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.

The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.

The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them. 

Examples

Input

2

Output

4
NNYY
NNYY
YYNN
YYNN

Input

9

Output

8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN

Input

1

Output

2
NY
YN

Note

In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.

In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
#include <bits/stdc++.h>
using namespace std;
int a[2002][2002];
int lay[500];
int k, n, t, st, now, i, p, j;
int main() {
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  n = 1;
  while (n <= k) {
    n *= 2;
    t++;
  }
  t--;
  n /= 2;
  st = 3;
  now = 2 * t;
  a[1][3] = 1;
  a[1][4] = 1;
  a[2][now + st - 2] = 1;
  a[2][now + st - 1] = 1;
  lay[1] = 4;
  int ej = 1;
  for (i = st; i <= st + now - 3; i += 2) {
    ej++;
    a[i][i + 2] = 1;
    a[i][i + 3] = 1;
    a[i + 1][i + 2] = 1;
    a[i + 1][i + 3] = 1;
    lay[ej] = i + 3;
  }
  k -= n;
  st += now;
  while (k > 0) {
    n = 1;
    p = 0;
    while (n <= k) {
      n *= 2;
      p++;
    }
    n /= 2;
    p--;
    if (p != t - 1) {
      a[1][st] = 1;
      for (i = 1; i < (t - p) - 1; i++) {
        a[st][st + 1] = 1;
        st++;
      }
      a[st][lay[t - p]] = 1;
      st++;
    } else {
      a[1][st] = 1;
      a[1][st + 1] = 1;
      for (i = 1; i < p; i++) {
        a[st][st + 2] = 1;
        a[st][st + 3] = 1;
        a[st + 1][st + 2] = 1;
        a[st + 1][st + 3] = 1;
        st += 2;
      }
      a[st][st + 2] = 1;
      a[st + 1][st + 2] = 1;
      st += 2;
      for (i = 1; i < t - p; i++) {
        a[st][st + 1] = 1;
        st++;
      }
      a[2][st] = 1;
      st++;
    }
    k -= n;
  }
  cout << st - 1 << endl;
  for (i = 1; i < st; i++) {
    for (j = 1; j < st; j++)
      if (a[i][j] || a[j][i])
        cout << "Y";
      else
        cout << "N";
    cout << endl;
  }
}
