#include <bits/stdc++.h>
using namespace std;
char G[1000][1001];
vector<int> adj[1000];
int cnt = 2;
void init(int lg, int c = 0) {
  int l = 0, r = 0;
  for (int i = (1); i <= (lg); i++) {
    adj[l].push_back(cnt);
    adj[l].push_back(cnt + 1);
    if (r) {
      adj[r].push_back(cnt);
      adj[r].push_back(cnt + 1);
    }
    l = cnt;
    r = cnt + 1;
    cnt += 2;
  }
  if (c) {
    adj[l].push_back(cnt);
    adj[r].push_back(cnt + 1);
    l = cnt;
    r = cnt + 1;
    cnt += 2;
  }
  adj[l].push_back(1);
  adj[r].push_back(1);
}
void append(int lg, int L) {
  int l = 2 * lg + 2;
  while (lg < L - 1) {
    adj[l].push_back(cnt);
    l = cnt++;
    lg++;
  }
  adj[l].push_back(1);
}
int main() {
  int k;
  cin >> k;
  if (k == 1) {
    puts("2\nNY\nYN");
    return 0;
  }
  int L;
  for (L = 31; L >= 0; L--)
    if (k & (1 << L)) break;
  init(L);
  for (int lg = L - 1; lg >= 0; lg--)
    if (k & (1 << lg)) {
      if (lg == L - 1)
        init(lg, 1);
      else
        append(lg, L);
    }
  printf("%d\n", cnt);
  for (int i = 0; i < (cnt); i++)
    for (int j = 0; j < (cnt); j++) G[i][j] = 'N';
  for (int u = 0; u < (cnt); u++)
    for (int v : adj[u]) G[u][v] = G[v][u] = 'Y';
  for (int u = 0; u < (cnt); u++) puts(G[u]);
  return 0;
}

### Prompt

Create a solution in CPP for the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char G[1000][1001];
vector<int> adj[1000];
int cnt = 2;
void init(int lg, int c = 0) {
  int l = 0, r = 0;
  for (int i = (1); i <= (lg); i++) {
    adj[l].push_back(cnt);
    adj[l].push_back(cnt + 1);
    if (r) {
      adj[r].push_back(cnt);
      adj[r].push_back(cnt + 1);
    }
    l = cnt;
    r = cnt + 1;
    cnt += 2;
  }
  if (c) {
    adj[l].push_back(cnt);
    adj[r].push_back(cnt + 1);
    l = cnt;
    r = cnt + 1;
    cnt += 2;
  }
  adj[l].push_back(1);
  adj[r].push_back(1);
}
void append(int lg, int L) {
  int l = 2 * lg + 2;
  while (lg < L - 1) {
    adj[l].push_back(cnt);
    l = cnt++;
    lg++;
  }
  adj[l].push_back(1);
}
int main() {
  int k;
  cin >> k;
  if (k == 1) {
    puts("2\nNY\nYN");
    return 0;
  }
  int L;
  for (L = 31; L >= 0; L--)
    if (k & (1 << L)) break;
  init(L);
  for (int lg = L - 1; lg >= 0; lg--)
    if (k & (1 << lg)) {
      if (lg == L - 1)
        init(lg, 1);
      else
        append(lg, L);
    }
  printf("%d\n", cnt);
  for (int i = 0; i < (cnt); i++)
    for (int j = 0; j < (cnt); j++) G[i][j] = 'N';
  for (int u = 0; u < (cnt); u++)
    for (int v : adj[u]) G[u][v] = G[v][u] = 'Y';
  for (int u = 0; u < (cnt); u++) puts(G[u]);
  return 0;
}

```