#include <bits/stdc++.h>
using namespace std;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double eps = 1e-10;
const double pi = acos(-1.0);
const long long INF = 1LL << 62;
const int inf = 1 << 29;
int k;
char graph[110][110];
void setting(int f, int t) {
  graph[f][t] = graph[t][f] = 'Y';
  return;
}
const int n = 95;
int main(void) {
  cin >> k;
  for (int i = (0); i < (110); ++i)
    for (int j = (0); j < (110); ++j) graph[i][j] = 'N';
  setting(1, 3);
  setting(1, 4);
  for (int i = (0); i < (30); ++i) {
    int f1 = 2 * i + 3, f2 = 2 * i + 4;
    int t1 = 2 * i + 5, t2 = 2 * i + 6;
    setting(f1, t1);
    setting(f2, t1);
    setting(f1, t2);
    setting(f2, t2);
  }
  for (int i = (0); i < (30); ++i) {
    int f = 65 + i, t = 65 + i + 1;
    setting(f, t);
  }
  setting(95, 2);
  for (int i = (0); i < (31); ++i) {
    int f = 2 * i + 4;
    int t = 65 + i;
    if (k & (1 << i)) setting(f, t);
  }
  cout << n << endl;
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = (1); j < (n + 1); ++j) cout << graph[i][j];
    cout << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double eps = 1e-10;
const double pi = acos(-1.0);
const long long INF = 1LL << 62;
const int inf = 1 << 29;
int k;
char graph[110][110];
void setting(int f, int t) {
  graph[f][t] = graph[t][f] = 'Y';
  return;
}
const int n = 95;
int main(void) {
  cin >> k;
  for (int i = (0); i < (110); ++i)
    for (int j = (0); j < (110); ++j) graph[i][j] = 'N';
  setting(1, 3);
  setting(1, 4);
  for (int i = (0); i < (30); ++i) {
    int f1 = 2 * i + 3, f2 = 2 * i + 4;
    int t1 = 2 * i + 5, t2 = 2 * i + 6;
    setting(f1, t1);
    setting(f2, t1);
    setting(f1, t2);
    setting(f2, t2);
  }
  for (int i = (0); i < (30); ++i) {
    int f = 65 + i, t = 65 + i + 1;
    setting(f, t);
  }
  setting(95, 2);
  for (int i = (0); i < (31); ++i) {
    int f = 2 * i + 4;
    int t = 65 + i;
    if (k & (1 << i)) setting(f, t);
  }
  cout << n << endl;
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = (1); j < (n + 1); ++j) cout << graph[i][j];
    cout << endl;
  }
  return 0;
}

```