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 n, st = 1, dr = 1, l_cur, pas;
char a[2005][2005];
int nod[2005], len[2005], kkt;
inline void Unite(int x1, int y1, int x2, int y2) {
  int i, j;
  for (i = x1; i <= y1; ++i)
    for (j = x2; j <= y2; ++j) a[i][j] = a[j][i] = 'Y';
}
inline void Solve(int x) {
  if (x == 1) {
    Unite(st, dr, 2, 2);
    ++l_cur;
    return;
  }
  int k = 0, i;
  while (x % 2 == 0) {
    Unite(st, dr, n + 1, n + 2);
    st = n + 1;
    dr = n + 2;
    ++l_cur;
    n += 2;
    x >>= 1;
  }
  if (x == 1) {
    Unite(st, dr, 2, 2);
    ++l_cur;
    return;
  }
  nod[++kkt] = ++n;
  len[kkt] = l_cur + 1;
  Unite(st, dr, n, n);
  Solve(x - 1);
}
int main() {
  int i, j, k;
  cin.sync_with_stdio(0);
  cin >> k;
  for (i = 1; i <= 1000; ++i)
    for (j = 1; j <= 1000; ++j) a[i][j] = 'N';
  n = 2;
  Solve(k);
  for (i = 1; i <= kkt; ++i) {
    int aux = nod[i];
    for (j = 1; j < l_cur - len[i]; ++j) {
      ++n;
      Unite(aux, aux, n, n);
      aux = n;
    }
    if (l_cur - len[i] > 1)
      Unite(n, n, 2, 2);
    else
      Unite(nod[i], nod[i], 2, 2);
  }
  cout << n << "\n";
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= n; ++j) cout << a[i][j];
    cout << "\n";
  }
  return 0;
}
