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>
bool adj_mat[(1000)][(1000)];
int n;
void add_edge(int i, int j) {
  adj_mat[i][j] = true;
  adj_mat[j][i] = true;
}
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 2; i <= 29; i++) {
    add_edge(i, i + 1);
  }
  add_edge(30, 1);
  int next_vertex = 31;
  for (int i = 0; i < 30; i++, k >>= 1) {
    assert(next_vertex + 2 * i <= (1000));
    if (!(k & 1)) continue;
    if (i == 0) {
      add_edge(0, 2);
      continue;
    }
    add_edge(0, next_vertex);
    add_edge(0, next_vertex + 1);
    for (int j = 0; j < 2 * i - 2; j += 2) {
      add_edge(next_vertex + j, next_vertex + j + 2);
      add_edge(next_vertex + j, next_vertex + j + 3);
      add_edge(next_vertex + j + 1, next_vertex + j + 2);
      add_edge(next_vertex + j + 1, next_vertex + j + 3);
    }
    if (i == 29) {
      add_edge(next_vertex + 2 * i - 2, 1);
      add_edge(next_vertex + 2 * i - 1, 1);
    } else {
      add_edge(next_vertex + 2 * i - 2, i + 2);
      add_edge(next_vertex + 2 * i - 1, i + 2);
    }
    next_vertex = next_vertex + 2 * i;
  }
  assert(next_vertex <= (1000));
  int n = next_vertex;
  printf("%d\n", n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      putchar(adj_mat[i][j] ? 'Y' : 'N');
    }
    printf("\n");
  }
  return 0;
}
