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>
#pragma warning(disable : 4996)
using namespace std;
int n, m, l[40][2], r[40];
int mp[1111][1111];
int main() {
  int i, j, k, x, y, z, ca = 1, re = 0;
  scanf("%d", &(n));
  for (int i = (int)(1); i <= (int)(200); ++i)
    for (int j = (int)(1); j <= (int)(200); ++j) mp[i][j] = 'N';
  int now = 3;
  r[0] = l[0][0] = l[0][1] = 1;
  r[30] = l[30][0] = l[30][1] = 2;
  for (int i = (int)(1); i <= (int)(29); ++i) r[i] = now++;
  for (int j = (int)(0); j <= (int)(28); ++j) mp[r[j]][r[j + 1]] = 'Y';
  for (int i = (int)(1); i <= (int)(29); ++i) l[i][0] = now++, l[i][1] = now++;
  for (int i = (int)(1); i <= (int)(29); ++i) {
    for (int j = 0; j < (int)(2); ++j)
      for (int k = 0; k < (int)(2); ++k) mp[l[i][j]][l[i + 1][k]] = 'Y';
  }
  for (int i = 0; i < (int)(30); ++i) {
    if (n & (1 << i)) {
      mp[r[29 - i]][l[30 - i][0]] = mp[r[29 - i]][l[30 - i][1]] = 'Y';
    }
  }
  n = 100;
  for (int i = (int)(1); i <= (int)(n); ++i)
    for (int j = (int)(1); j <= (int)(n); ++j)
      if (mp[i][j] == 'Y' || mp[j][i] == 'Y') mp[i][j] = mp[j][i] = 'Y';
  printf("%d\n", (100));
  for (int i = (int)(1); i <= (int)(n); ++i) {
    for (int j = (int)(1); j <= (int)(n); ++j) putchar(mp[i][j]);
    putchar('\n');
  }
  return 0;
}