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 k;
char mat[1200][1200];
int n;
void add(int u, int v) {
  mat[u][v] = 'Y';
  mat[v][u] = 'Y';
}
int arr[100];
int main() {
  scanf("%d", &k);
  int cnt = 0;
  while (k) {
    arr[cnt++] = k % 10;
    k /= 10;
  }
  int n = 3;
  int st, ed, tst, ted;
  for (int i = cnt - 1; i >= 0; --i) {
    if (!arr[i]) continue;
    st = 1, ed = 1;
    for (int j = st; j <= ed; j++) {
      for (int k = 0; k < arr[i]; k++) add(j, n + k);
    }
    st = n, ed = n + arr[i] - 1;
    n += arr[i];
    for (int j = 1; j <= i; j++) {
      for (k = st; k <= ed; k++) {
        for (int l = 0; l < 10; l++) add(k, n + l);
      }
      st = n, ed = n + 10 - 1;
      n += 10;
    }
    for (int j = i + 1; j <= cnt - 1; j++) {
      for (k = st; k <= ed; k++) add(k, n);
      st = n, ed = n;
      n++;
    }
    for (int j = st; j <= ed; j++) add(j, 2);
  }
  for (int i = 1; i <= n - 1; i++) {
    mat[i][n] = '\0';
    for (int j = 1; j <= n - 1; j++)
      if (!mat[i][j]) mat[i][j] = 'N';
  }
  printf("%d\n", n - 1);
  for (int i = 1; i <= n - 1; i++) printf("%s\n", mat[i] + 1);
  return 0;
}