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;
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;
const long long linf = 0x3f3f3f3f3f3f3f3f;
bool mat[1010][1010];
int main() {
  int k;
  scanf("%d", &k);
  k--;
  if (k == 0) {
    return 0 * printf("2\nNY\nYN\n");
  }
  int l = 12, m = 33;
  int gr = 0;
  mat[l - 1][l] = true;
  for (int i = 0; i < 31; i++) {
    if (k & (1 << i)) gr = i;
    mat[l + i][l + i + 1] = true;
  }
  mat[l + m - 1][l + m] = true;
  for (int i = 0; i < gr - 1; i++) {
    mat[l + i + m][l + i + m + 1] = true;
    mat[l + i + m][l + i + 1] = true;
    mat[l + i][l + i + m + 1] = true;
  }
  int n = l + m + gr;
  if (gr > 0) mat[l + m + gr - 1][l + gr] = true, n++;
  mat[n - 1][n] = true;
  for (int i = 0; i < 31; i++) {
    mat[n + i][n + i + 1] = true;
  }
  for (int i = 0; i < gr; i++) {
    if (k & (1 << i)) mat[n + gr - i - 2][l + gr - i - 1] = true;
  }
  mat[1][l - 1] = true;
  mat[1][l + m - 1] = true;
  mat[1][n - 1] = true;
  mat[2][l + m - 2] = true;
  mat[2][n + 31] = true;
  int p = n + 31 + 10;
  printf("%d\n", p);
  for (int i = 1; i <= p; i++) {
    for (int j = 1; j <= p; j++) {
      printf("%c", (mat[i][j] || mat[j][i]) ? 'Y' : 'N');
    }
    printf("\n");
  }
  return 0;
}
