#include <bits/stdc++.h>
using namespace std;
const int N = 1111;
int node_cnt;
char mat[N][N];
vector<int> dfs(int n) {
  vector<int> res;
  if (n == 1) {
    res.push_back(0);
    res.push_back(1);
    return res;
  }
  vector<int> last = dfs(n >> 1);
  int cnt = last[0] + 1;
  res.push_back(cnt);
  for (int d = 0; d < 2; d++) {
    res.push_back(++node_cnt);
    for (int i = 1, sz = last.size(); i < sz; i++) {
      mat[node_cnt][last[i]] = mat[last[i]][node_cnt] = 'Y';
    }
  }
  if (n & 1) {
    node_cnt++;
    mat[1][node_cnt] = mat[node_cnt][1] = 'Y';
    for (int i = 1; i < cnt; i++, node_cnt++)
      mat[node_cnt][node_cnt + 1] = mat[node_cnt + 1][node_cnt] = 'Y';
    res.push_back(node_cnt);
  }
  return res;
}
void build(int n) {
  node_cnt = 2;
  memset(mat, 0, sizeof mat);
  vector<int> pv = dfs(n);
  for (int i = 1, sz = pv.size(); i < sz; i++) {
    mat[pv[i]][2] = mat[2][pv[i]] = 'Y';
  }
  for (int i = 1; i <= node_cnt; i++) {
    for (int j = 1; j <= node_cnt; j++) {
      if (mat[i][j]) continue;
      mat[i][j] = 'N';
    }
  }
}
int main() {
  int n;
  while (cin >> n) {
    build(n);
    cout << node_cnt << endl;
    for (int i = 1; i <= node_cnt; i++) cout << mat[i] + 1 << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int N = 1111;
int node_cnt;
char mat[N][N];
vector<int> dfs(int n) {
  vector<int> res;
  if (n == 1) {
    res.push_back(0);
    res.push_back(1);
    return res;
  }
  vector<int> last = dfs(n >> 1);
  int cnt = last[0] + 1;
  res.push_back(cnt);
  for (int d = 0; d < 2; d++) {
    res.push_back(++node_cnt);
    for (int i = 1, sz = last.size(); i < sz; i++) {
      mat[node_cnt][last[i]] = mat[last[i]][node_cnt] = 'Y';
    }
  }
  if (n & 1) {
    node_cnt++;
    mat[1][node_cnt] = mat[node_cnt][1] = 'Y';
    for (int i = 1; i < cnt; i++, node_cnt++)
      mat[node_cnt][node_cnt + 1] = mat[node_cnt + 1][node_cnt] = 'Y';
    res.push_back(node_cnt);
  }
  return res;
}
void build(int n) {
  node_cnt = 2;
  memset(mat, 0, sizeof mat);
  vector<int> pv = dfs(n);
  for (int i = 1, sz = pv.size(); i < sz; i++) {
    mat[pv[i]][2] = mat[2][pv[i]] = 'Y';
  }
  for (int i = 1; i <= node_cnt; i++) {
    for (int j = 1; j <= node_cnt; j++) {
      if (mat[i][j]) continue;
      mat[i][j] = 'N';
    }
  }
}
int main() {
  int n;
  while (cin >> n) {
    build(n);
    cout << node_cnt << endl;
    for (int i = 1; i <= node_cnt; i++) cout << mat[i] + 1 << endl;
  }
  return 0;
}

```