Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, k;
bool matr[2005][2005];
int main() {
  scanf("%d", &k);
  n = 2;
  int len = 0;
  for (int i = 1, deg = 0; i <= k; i = (i << 1), deg++) {
    if (i & k) len = deg;
  }
  for (int deg = len; deg >= 0; deg--) {
    if (!(k & (1 << deg))) continue;
    if (len > deg) {
      n++;
      matr[1 + 2 * deg][n] = matr[n][1 + 2 * deg] = true;
      for (int j = 0; j < len - deg; j++) {
        n++;
        matr[n][n - 1] = matr[n - 1][n] = true;
      }
      matr[2][n] = matr[n][2] = true;
      n++;
      matr[2 + 2 * deg][n] = matr[n][2 + 2 * deg] = true;
      for (int j = 0; j < len - deg; j++) {
        n++;
        matr[n][n - 1] = matr[n - 1][n] = true;
      }
      matr[2][n] = matr[n][2] = true;
      continue;
    }
    if (deg == 0) {
      if (len >= 1) {
        n++;
        matr[n][1] = matr[1][n] = true;
        for (int j = 0; j < len - 1; j++) {
          n++;
          matr[n][n - 1] = matr[n - 1][n] = true;
        }
        matr[n][2] = matr[2][n] = true;
      } else
        matr[1][2] = matr[2][1] = true;
      continue;
    }
    for (int j = 0; j < 2; j++) {
      n++;
      matr[1][n] = matr[n][1] = true;
    }
    for (int j = 2; j <= deg; j++) {
      n++;
      matr[n - 2][n] = matr[n][n - 2] = true;
      matr[n - 1][n] = matr[n][n - 1] = true;
      n++;
      matr[n - 3][n] = matr[n][n - 3] = true;
      matr[n - 2][n] = matr[n][n - 2] = true;
    }
    for (int j = deg; j <= len; j++) {
      n++;
      matr[n][n - 2] = matr[n - 2][n] = true;
      n++;
      matr[n][n - 2] = matr[n - 2][n] = true;
    }
    matr[n][2] = matr[2][n] = true;
    matr[n - 1][2] = matr[2][n - 1] = true;
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (matr[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}