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[500005];
bool v[1001][1001];
int b[500005][2];
int c[500005];
int n, m, tot;
void add(int x, int y) {
  v[x][y] = 1;
  v[y][x] = 1;
}
void Solve() {
  int i, j, t, len, x, y, tx, ty, last;
  cin >> n;
  memset(v, 0, sizeof(v));
  a[1] = 1;
  m = 31;
  for (i = 2; i <= m; ++i) a[i] = a[i - 1] * 2;
  x = 3;
  y = 4;
  b[m][0] = x;
  b[m][1] = y;
  x = 5;
  y = 6;
  for (i = m - 1; i >= 2; --i) {
    add(b[i + 1][0], x);
    add(b[i + 1][1], x);
    add(b[i + 1][0], y);
    add(b[i + 1][1], y);
    b[i][0] = x;
    b[i][1] = y;
    x += 2;
    y += 2;
  }
  add(b[2][0], x);
  add(b[2][1], x);
  b[1][0] = b[1][1] = x;
  add(2, x);
  ++x;
  for (i = 1; i <= m; ++i) add(x, x + 1), c[i] = x++;
  for (i = m; i >= 1; --i) {
    if (n >= a[i]) {
      n -= a[i];
      t = m - i;
      x = 1;
      if (t > 0) {
        x = c[t];
        add(1, c[1]);
      }
      add(x, b[i][0]);
      add(x, b[i][1]);
    }
  }
  cout << 100 << endl;
  for (i = 1; i <= 100; ++i) {
    for (j = 1; j <= 100; ++j)
      if (v[i][j])
        cout << "Y";
      else
        cout << "N";
    cout << endl;
  }
}
int main() {
  Solve();
  cin >> n;
  return 0;
}
