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;
vector<int> e[1000];
int a[100];
int main() {
  int k;
  scanf("%d", &k);
  if (k == 1) {
    printf("2\nNY\nYN");
    return 0;
  }
  int timer = 2, len = 0;
  for (int cur = 0, i = 1; k >= (1 << i); i++) {
    e[cur].push_back(timer);
    e[cur].push_back(timer + 1);
    e[timer].push_back(timer + 2);
    e[timer + 1].push_back(timer + 2);
    cur = timer + 2;
    timer += 3;
    a[i] = cur;
    len += 2;
  }
  e[timer - 1].push_back(1);
  len++;
  char flag = false;
  for (int i = 30; i >= 0; i--) {
    if ((k & (1 << i)) == 0) continue;
    if (!flag) {
      flag = true;
      continue;
    }
    int cur_len = i * 2, cur = a[i];
    while (cur_len + 1 < len) {
      e[cur].push_back(timer);
      cur = timer++;
      cur_len++;
    }
    e[cur].push_back(1);
  }
  int n = timer;
  char s[n][n];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) s[i][j] = 'N';
  for (int v = 0; v < n; v++)
    for (int to : e[v]) s[v][to] = s[to][v] = 'Y';
  printf("%d\n", n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) printf("%c", s[i][j]);
    printf("\n");
  }
  return 0;
}