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 k;
char m[1005][1005];
int alloc = 3;
void add_edge(int a, int b) {
  --a;
  --b;
  m[a][b] = m[b][a] = 'Y';
}
void attach(int n, int cnt) {
  vector<int> add;
  for (int i = (0); i < (cnt); ++i) {
    add.push_back(alloc);
    add_edge(n, alloc);
    ++alloc;
  }
  for (int i = (0); i < (cnt); ++i) add_edge(add[i], alloc);
  ++alloc;
}
int main(void) {
  scanf("%d", &k);
  int base = 8;
  int top = 10;
  for (int i = top; i >= 0; --i) {
    int q = ((k >> (3 * i)) % base);
    if (q == 0) continue;
    attach(1, q);
    for (int j = (0); j < (i); ++j) attach(alloc - 1, 8);
    for (int j = (0); j < (top - i); ++j) attach(alloc - 1, 1);
    add_edge(alloc - 1, 2);
  }
  int n = alloc - 1;
  printf("%d\n", n);
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (n); ++j)
      if (!m[i][j]) m[i][j] = 'N';
  for (int i = (0); i < (n); ++i) puts(m[i]);
  return 0;
}
