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 n, nnode;
char g[1010][1010];
vector<int> dfs(int k) {
  int i, j, len;
  vector<int> res, befo;
  if (k == 1) {
    res.push_back(0);
    res.push_back(1);
    return res;
  }
  befo = dfs(k / 2);
  len = befo[0] + 1;
  res.push_back(len);
  for (i = 0; i < 2; i++) {
    res.push_back(++nnode);
    for (j = 1; j < befo.size(); j++)
      g[nnode][befo[j]] = g[befo[j]][nnode] = 'Y';
  }
  if (k & 1) {
    ++nnode;
    g[1][nnode] = g[nnode][1] = 'Y';
    for (i = 1; i < len; i++) {
      g[nnode][nnode + 1] = g[nnode + 1][nnode] = 'Y';
      nnode++;
    }
    res.push_back(nnode);
  }
  return res;
}
void solve() {
  int i, j;
  vector<int> res;
  memset(g, 0, sizeof(g));
  nnode = 2;
  res = dfs(n);
  for (i = 1; i < res.size(); i++) g[2][res[i]] = g[res[i]][2] = 'Y';
  for (i = 1; i <= nnode; i++)
    for (j = 1; j <= nnode; j++)
      if (g[i][j] == 0) g[i][j] = 'N';
  printf("%d\n", nnode);
  for (i = 1; i <= nnode; i++) {
    for (j = 1; j <= nnode; j++) printf("%c", g[i][j]);
    printf("\n");
  }
}
void input() {
  scanf("%d", &n);
  solve();
}
int main() {
  input();
  fclose(stdin);
  fclose(stdout);
  return 0;
}