#include <bits/stdc++.h>
using namespace std;
int vcnt = 2;
bool edge[1010][1010];
vector<int> v[16 + 10];
int b[16 + 10];
long long p4[16 + 10];
void makePart() {
  v[0].push_back(0);
  for (int i = 0; i < 16; i++) {
    for (int j = 0; j < b[i]; j++) {
      for (int x = 0; x < v[i].size(); x++)
        edge[v[i][x]][vcnt] = edge[vcnt][v[i][x]] = true;
      v[i + 1].push_back(vcnt);
      vcnt++;
    }
    v[i].clear();
  }
  for (int x = 0; x < v[16].size(); x++)
    edge[v[16][x]][1] = edge[1][v[16][x]] = true;
  v[16].clear();
}
int main() {
  int n;
  cin >> n;
  p4[0] = 1;
  for (int i = 1; i < 16; i++) p4[i] = p4[i - 1] * 4;
  for (int i = 16 - 2; i >= 0; i--) {
    if (n >= p4[i]) {
      for (int j = 0; j < 16 - 1; j++)
        if (j < i)
          b[j] = 4;
        else
          b[j] = 1;
      b[16 - 1] = n / p4[i];
      makePart();
      n %= p4[i];
    }
  }
  cout << vcnt << endl;
  for (int i = 0; i < vcnt; i++) {
    for (int j = 0; j < vcnt; j++)
      if (!edge[i][j])
        cout << 'N';
      else
        cout << 'Y';
    cout << endl;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 vcnt = 2;
bool edge[1010][1010];
vector<int> v[16 + 10];
int b[16 + 10];
long long p4[16 + 10];
void makePart() {
  v[0].push_back(0);
  for (int i = 0; i < 16; i++) {
    for (int j = 0; j < b[i]; j++) {
      for (int x = 0; x < v[i].size(); x++)
        edge[v[i][x]][vcnt] = edge[vcnt][v[i][x]] = true;
      v[i + 1].push_back(vcnt);
      vcnt++;
    }
    v[i].clear();
  }
  for (int x = 0; x < v[16].size(); x++)
    edge[v[16][x]][1] = edge[1][v[16][x]] = true;
  v[16].clear();
}
int main() {
  int n;
  cin >> n;
  p4[0] = 1;
  for (int i = 1; i < 16; i++) p4[i] = p4[i - 1] * 4;
  for (int i = 16 - 2; i >= 0; i--) {
    if (n >= p4[i]) {
      for (int j = 0; j < 16 - 1; j++)
        if (j < i)
          b[j] = 4;
        else
          b[j] = 1;
      b[16 - 1] = n / p4[i];
      makePart();
      n %= p4[i];
    }
  }
  cout << vcnt << endl;
  for (int i = 0; i < vcnt; i++) {
    for (int j = 0; j < vcnt; j++)
      if (!edge[i][j])
        cout << 'N';
      else
        cout << 'Y';
    cout << endl;
  }
}

```