#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 5;
int n;
int S, E;
int RS, RE;
char mp[N][N];
vector<int> v;
void Init() { memset(mp, 'N', sizeof(mp)); }
void Calc() {
  int k = n;
  v.clear();
  for (int i = 0; i < 32; i++) {
    if (k & (1 << i)) {
      v.push_back(i);
    }
  }
  reverse(v.begin(), v.end());
}
void Solve() {
  mp[1][3] = mp[3][1] = 'Y';
  int pre = 3;
  int now;
  S = 3;
  for (int i = 1; i <= v[0]; i++) {
    for (int j = 1; j <= 2; j++) {
      mp[pre][pre + j] = mp[pre + j][pre] = 'Y';
    }
    now = pre + 3;
    for (int j = 1; j <= 2; j++) {
      mp[pre + j][now] = mp[now][pre + j] = 'Y';
    }
    pre = now;
  }
  E = pre;
  mp[E][2] = mp[2][E] = 'Y';
}
void Build() {
  if ((int)v.size() == 1) {
    RE = E;
    return;
  }
  RS = E + 1;
  RE = RS + 2 * (v[0] - 1);
  for (int i = RS; i < RE; i++) {
    mp[i][i + 1] = mp[i + 1][i] = 'Y';
  }
  mp[RE][E] = mp[E][RE] = 'Y';
  int len = (int)v.size();
  for (int i = 1; i < len; i++) {
    int s = S + 3 * v[i];
    int e = RS + 2 * v[i];
    mp[s][e] = mp[e][s] = 'Y';
  }
}
void Print() {
  printf("%d\n", RE);
  for (int i = 1; i <= RE; i++) {
    for (int j = 1; j <= RE; j++) {
      printf("%c", mp[i][j]);
    }
    puts("");
  }
}
int main() {
  while (~scanf("%d", &n)) {
    Init();
    Calc();
    Solve();
    Build();
    Print();
  }
  return 0;
}

### 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;
const int N = 1000 + 5;
int n;
int S, E;
int RS, RE;
char mp[N][N];
vector<int> v;
void Init() { memset(mp, 'N', sizeof(mp)); }
void Calc() {
  int k = n;
  v.clear();
  for (int i = 0; i < 32; i++) {
    if (k & (1 << i)) {
      v.push_back(i);
    }
  }
  reverse(v.begin(), v.end());
}
void Solve() {
  mp[1][3] = mp[3][1] = 'Y';
  int pre = 3;
  int now;
  S = 3;
  for (int i = 1; i <= v[0]; i++) {
    for (int j = 1; j <= 2; j++) {
      mp[pre][pre + j] = mp[pre + j][pre] = 'Y';
    }
    now = pre + 3;
    for (int j = 1; j <= 2; j++) {
      mp[pre + j][now] = mp[now][pre + j] = 'Y';
    }
    pre = now;
  }
  E = pre;
  mp[E][2] = mp[2][E] = 'Y';
}
void Build() {
  if ((int)v.size() == 1) {
    RE = E;
    return;
  }
  RS = E + 1;
  RE = RS + 2 * (v[0] - 1);
  for (int i = RS; i < RE; i++) {
    mp[i][i + 1] = mp[i + 1][i] = 'Y';
  }
  mp[RE][E] = mp[E][RE] = 'Y';
  int len = (int)v.size();
  for (int i = 1; i < len; i++) {
    int s = S + 3 * v[i];
    int e = RS + 2 * v[i];
    mp[s][e] = mp[e][s] = 'Y';
  }
}
void Print() {
  printf("%d\n", RE);
  for (int i = 1; i <= RE; i++) {
    for (int j = 1; j <= RE; j++) {
      printf("%c", mp[i][j]);
    }
    puts("");
  }
}
int main() {
  while (~scanf("%d", &n)) {
    Init();
    Calc();
    Solve();
    Build();
    Print();
  }
  return 0;
}

```