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>
char con[1000][1000];
int main() {
  int n;
  scanf("%d", &n);
  int h = -1;
  for (int i = 30; i >= 0; i--) {
    if ((1 << i) & n) {
      if (h == -1) h = i;
    }
  }
  if (n == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  int t = 2;
  int x = 0;
  for (int j = 0; j < h - 1; j++) {
    con[x][t] = con[t][x] = 'Y';
    con[x][t + 1] = con[t + 1][x] = 'Y';
    con[t][t + 2] = con[t + 2][t] = 'Y';
    con[t + 1][t + 2] = con[t + 2][t + 1] = 'Y';
    x = t + 2;
    t += 3;
  }
  con[x][t] = con[t][x] = 'Y';
  con[x][t + 1] = con[t + 1][x] = 'Y';
  con[t][1] = con[1][t] = 'Y';
  con[t + 1][1] = con[1][t + 1] = 'Y';
  t += 2;
  for (int i = h - 1; i >= 0; i--) {
    if ((1 << i) & n) {
      int x = i == 0 ? 0 : (3 * i + 1);
      for (int j = i; j < h - 1; j++) {
        con[x][t] = con[t][x] = 'Y';
        con[t][t + 1] = con[t + 1][t] = 'Y';
        x = t + 1;
        t += 2;
      }
      con[x][t] = con[t][x] = 'Y';
      con[t][1] = con[1][t] = 'Y';
      t += 1;
    }
  }
  printf("%d\n", t);
  for (int i = 0; i < t; i++) {
    for (int j = 0; j < t; j++) {
      if (con[i][j] == 'Y')
        printf("Y");
      else
        printf("N");
    }
    puts("");
  }
}