#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> adjMat(1001, vector<int>(1001));
void connect(vector<int> &a, vector<int> &b) {
  for (int x : a) {
    for (int y : b) adjMat[x][y] = adjMat[y][x] = 1;
  }
  a = move(b);
  b.clear();
}
int main() {
  int k;
  cin >> k;
  int node = 3;
  for (int i = 0; i < 15; i++) {
    int curr = k % 4;
    k -= k % 4;
    k /= 4;
    if (curr == 0) continue;
    vector<int> prev{1};
    vector<int> next;
    for (int j = 0; j < curr; j++) next.push_back(node++);
    connect(prev, next);
    int j = 1;
    for (; j <= i; j++) {
      for (int k = 0; k < 4; k++) next.push_back(node++);
      connect(prev, next);
    }
    for (; j < 15; j++) {
      next.push_back(node++);
      connect(prev, next);
    }
    next.push_back(2);
    connect(prev, next);
  }
  cout << node - 1 << endl;
  for (int i = 1; i < node; i++) {
    for (int j = 1; j < node; j++) printf("%c", adjMat[i][j] == 1 ? 'Y' : 'N');
    printf("\n");
  }
}

### Prompt

Your task is to create a CPP solution to 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;
vector<vector<int>> adjMat(1001, vector<int>(1001));
void connect(vector<int> &a, vector<int> &b) {
  for (int x : a) {
    for (int y : b) adjMat[x][y] = adjMat[y][x] = 1;
  }
  a = move(b);
  b.clear();
}
int main() {
  int k;
  cin >> k;
  int node = 3;
  for (int i = 0; i < 15; i++) {
    int curr = k % 4;
    k -= k % 4;
    k /= 4;
    if (curr == 0) continue;
    vector<int> prev{1};
    vector<int> next;
    for (int j = 0; j < curr; j++) next.push_back(node++);
    connect(prev, next);
    int j = 1;
    for (; j <= i; j++) {
      for (int k = 0; k < 4; k++) next.push_back(node++);
      connect(prev, next);
    }
    for (; j < 15; j++) {
      next.push_back(node++);
      connect(prev, next);
    }
    next.push_back(2);
    connect(prev, next);
  }
  cout << node - 1 << endl;
  for (int i = 1; i < node; i++) {
    for (int j = 1; j < node; j++) printf("%c", adjMat[i][j] == 1 ? 'Y' : 'N');
    printf("\n");
  }
}

```