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 ini, k, ck, sum, spath;
bool m[1024][1024];
vector<int> v[12], a[12], b[12];
int extra[] = {0, 0, 1, 1, 2, 1, 2, 1, 3, 2};
void add(int x, int p) {
  v[p].push_back(x);
  a[p].push_back(ini);
  b[p].push_back(ini + x - 1);
  ini = ini + x;
  sum += x;
}
void salva(int q, int p) {
  v[p].clear();
  a[p].clear();
  b[p].clear();
  spath = max(extra[q] + 2 * p, spath);
  if (p == 0 && q == 1) {
    add(1, p);
    return;
  }
  for (int i = 2; i <= q;) {
    if (q % i == 0) {
      add(i, p);
      q /= i;
    } else
      i++;
  }
  for (int i = 0; i < p; i++) {
    add(2, p);
    add(5, p);
  }
}
int main() {
  int p1, p2, powr;
  scanf(" %d", &k);
  if (k == 1) {
    printf("2\nNY\nYN\n");
    return 0;
  }
  ck = k;
  powr = spath = 0;
  ini = 3;
  sum = 2;
  while (ck > 0) {
    if (ck % 10) salva(ck % 10, powr);
    ck /= 10;
    powr++;
  }
  for (int i = 0; i < 1024; i++)
    for (int j = 0; j < 1024; j++) m[i][j] = false;
  for (int i = 0; i < powr; i++) {
    if (v[i].size() == 0) continue;
    while (v[i].size() < spath) {
      add(1, i);
    }
    p1 = p2 = 1;
    for (int cnt = 0; cnt < v[i].size(); cnt++) {
      for (int h = p1; h <= p2; h++) {
        for (int j = a[i][cnt]; j <= b[i][cnt]; j++) m[h][j] = m[j][h] = true;
      }
      if (cnt == v[i].size() - 1)
        for (int j = a[i][cnt]; j <= b[i][cnt]; j++) m[2][j] = m[j][2] = true;
      p1 = a[i][cnt];
      p2 = b[i][cnt];
    }
  }
  printf("%d\n", sum);
  for (int i = 1; i <= sum; i++) {
    for (int j = 1; j <= sum; j++) printf("%c", m[i][j] ? 'Y' : 'N');
    printf("\n");
  }
  return 0;
}