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>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
const int N = 1005;
const int INF = 1e9 + 7;
char s[N][N];
int n, G[N], D[N], v[N];
void Add(int x, int y) { s[x][y] = s[y][x] = 'Y'; }
void Build(int cnt, int t) {
  if (!t) {
    while (cnt--) {
      n++;
      Add(1, n);
      for (int i = 0; i < 10; i++) {
        Add(n, n + 1);
        n++;
      }
      Add(n, 2);
    }
    return;
  }
  n += 10;
  for (int i = 1; i <= 10; i++) Add(1, n - i + 1);
  for (int i = 0; i < t - 1; i++) {
    n += 10;
    for (int x = 1; x <= 10; x++)
      for (int y = 1; y <= 10; y++) Add(n - 20 + x, n - 10 + y);
  }
  for (int x = 1; x <= 10; x++)
    for (int y = 1; y <= cnt; y++) Add(n - 10 + x, n + y);
  for (int y = 1; y <= cnt; y++) Add(n + y, n + cnt + 1);
  n += cnt + 1;
  for (int i = t; i < 9; i++) {
    Add(n, n + 1);
    n++;
  }
  Add(n, 2);
}
int main() {
  int K;
  for (int i = 0; i < 1000; i++)
    for (int j = 0; j < 1000; j++) s[i][j] = s[j][i] = 'N';
  cin >> K;
  n = 2;
  int t = 0;
  for (int i = 1; i <= K; i *= 10) {
    int cnt = K % ((long long)i * 10) / i;
    Build(cnt, t);
    t++;
  }
  cout << n << endl;
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      printf("%c", s[i][j]);
    }
    puts("");
  }
}
