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 lim, val, dep;
int adj_mat[1009][1009];
void recurse(int i, int j, int k, int depth) {
  if (k & val) {
    lim = (j == 1 ? j + 2 : j + 1);
    for (int q = int(i); q <= int(j); q++) {
      adj_mat[q][lim] = 1;
      adj_mat[lim][q] = 1;
    }
    for (int q = int(depth); q <= int(dep - 1); q++) {
      adj_mat[lim][lim + 1] = 1;
      adj_mat[lim + 1][lim] = 1;
      lim++;
    }
    adj_mat[lim][2] = 1;
    adj_mat[2][lim] = 1;
  }
  if (dep == depth) return;
  int add = max((j == 1 ? 3 : j + 1), lim + 1);
  for (int p = int(i); p <= int(j); p++) {
    for (int x = int(add); x <= int(add + 1); x++) {
      adj_mat[p][x] = 1;
      adj_mat[x][p] = 1;
    }
  }
  recurse(add, add + 1, k * 2, depth + 1);
}
int main() {
  memset(adj_mat, 0, sizeof(adj_mat));
  val;
  scanf("%d", &val);
  lim = 2;
  dep = -1;
  for (int i = 31; i >= 0; i--) {
    if (val & (1 << i)) dep = max(dep, i);
  }
  recurse(1, 1, 1, 0);
  printf("%d\n", lim);
  for (int i = int(1); i <= int(lim); i++) {
    for (int j = int(1); j <= int(lim); j++) {
      if (adj_mat[i][j] == 1)
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}