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>
using namespace std;
int a[1010][1010], n;
void addPath(int pos, int digit) {
  int prev = 0, next;
  for (int i = 0; i < 10; i++) {
    int lim = 10;
    if (i == pos)
      lim = digit;
    else if (i > pos)
      lim = 1;
    next = n + lim;
    if (i == 9) next = 1;
    for (int j = 0; j < lim; j++) {
      a[prev][n + j] = a[n + j][prev] = 1;
      a[next][n + j] = a[n + j][next] = 1;
    }
    prev = next;
    if (i < 9)
      n = next + 1;
    else
      n++;
  }
}
int main() {
  ios::sync_with_stdio(0);
  int k;
  cin >> k;
  n = 2;
  for (int i = 0; k; i++) {
    addPath(i, k % 10);
    k /= 10;
  }
  cout << n << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cout << (a[i][j] ? 'Y' : 'N');
    cout << "\n";
  }
}
