#include <bits/stdc++.h>
using namespace std;
int k, n;
bool g[1234][1234];
bool prime = false;
vector<int> pf;
vector<int> v[1234];
void scan() { cin >> k; }
void fact(int xx) {
  int x = xx;
  for (int i = 2; i <= 300; i++) {
    while (x % i == 0) {
      pf.push_back(i);
      x /= i;
    }
  }
  if (x > 1) {
    pf.push_back(-1);
    fact(x - 1);
  }
}
void solve() {
  int z = pf.size();
  int cnt = 3;
  v[0].push_back(1);
  v[z + 1].push_back(2);
  for (int i = 0; i < z; i++) {
    if (pf[i] == -1) {
      v[i + 1].push_back(cnt);
      cnt++;
      continue;
    }
    for (int j = 0; j < pf[i]; j++) {
      v[i + 1].push_back(cnt++);
    }
  }
  int t = 0;
  for (int i = z - 1; i >= 0; i--) {
    t++;
    if (pf[i] != -1) continue;
    int xz = v[i + 1][0];
    int lst = xz;
    for (int j = 1; j < t; j++) {
      g[lst][cnt] = g[cnt][lst] = true;
      lst = cnt;
      cnt++;
    }
    g[2][lst] = g[lst][2] = true;
    cnt++;
  }
  n = cnt - 1;
}
void cons() {
  int z = pf.size();
  for (int i = 0; i <= z; i++) {
    for (int j = 0; j < v[i].size(); j++) {
      for (int jj = 0; jj < v[i + 1].size(); jj++) {
        g[v[i][j]][v[i + 1][jj]] = g[v[i + 1][jj]][v[i][j]] = true;
      }
    }
  }
}
void print() {
  int z = pf.size();
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (g[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
}
int main() {
  scan();
  fact(k);
  solve();
  cons();
  print();
  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;
int k, n;
bool g[1234][1234];
bool prime = false;
vector<int> pf;
vector<int> v[1234];
void scan() { cin >> k; }
void fact(int xx) {
  int x = xx;
  for (int i = 2; i <= 300; i++) {
    while (x % i == 0) {
      pf.push_back(i);
      x /= i;
    }
  }
  if (x > 1) {
    pf.push_back(-1);
    fact(x - 1);
  }
}
void solve() {
  int z = pf.size();
  int cnt = 3;
  v[0].push_back(1);
  v[z + 1].push_back(2);
  for (int i = 0; i < z; i++) {
    if (pf[i] == -1) {
      v[i + 1].push_back(cnt);
      cnt++;
      continue;
    }
    for (int j = 0; j < pf[i]; j++) {
      v[i + 1].push_back(cnt++);
    }
  }
  int t = 0;
  for (int i = z - 1; i >= 0; i--) {
    t++;
    if (pf[i] != -1) continue;
    int xz = v[i + 1][0];
    int lst = xz;
    for (int j = 1; j < t; j++) {
      g[lst][cnt] = g[cnt][lst] = true;
      lst = cnt;
      cnt++;
    }
    g[2][lst] = g[lst][2] = true;
    cnt++;
  }
  n = cnt - 1;
}
void cons() {
  int z = pf.size();
  for (int i = 0; i <= z; i++) {
    for (int j = 0; j < v[i].size(); j++) {
      for (int jj = 0; jj < v[i + 1].size(); jj++) {
        g[v[i][j]][v[i + 1][jj]] = g[v[i + 1][jj]][v[i][j]] = true;
      }
    }
  }
}
void print() {
  int z = pf.size();
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (g[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
}
int main() {
  scan();
  fact(k);
  solve();
  cons();
  print();
  return 0;
}

```