#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
bool grid[N][N];
long long k;
vector<pair<int, int> > edges;
int n = 1, depth = 1, pt = 3;
void generate_ans(int mx) {
  vector<int> p;
  p.push_back(1);
  for (int i = 0; i < mx; i++) {
    int idx = max(0, int(p.size()) - 2);
    for (int j = idx; j < p.size(); j++)
      grid[pt][p[j]] = grid[p[j]][pt] = grid[pt + 1][p[j]] =
          grid[p[j]][pt + 1] = 1;
    p.push_back(pt);
    p.push_back(pt + 1);
    if (i + 1 == depth)
      grid[pt][2] = grid[2][pt] = grid[pt + 1][2] = grid[2][pt + 1] = 1;
    pt += 2;
  }
  mx = depth - mx;
  if (mx) {
    int idx = max(0, int(p.size()) - 2);
    while (idx < p.size()) {
      int a = p[idx];
      grid[a][37 - mx] = grid[37 - mx][a] = 1;
      idx++;
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> k;
  if (k == 1) {
    cout << "2\nNY\nYN";
    return 0;
  }
  for (int i = 3; i < 36; i++) grid[i][i + 1] = grid[i + 1][i] = 1;
  grid[36][2] = grid[2][36] = 1;
  pt = 38;
  for (int i = 30; i >= 0; i--) {
    if (k & (1 << i)) depth = max(depth, i), generate_ans(i);
  }
  cout << pt - 1 << endl;
  char z[5] = "NY";
  for (int i = 1; i < pt; i++) {
    for (int j = 1; j < pt; j++) printf("%c", z[grid[i][j]]);
    printf("\n");
  }
  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 = 5005;
bool grid[N][N];
long long k;
vector<pair<int, int> > edges;
int n = 1, depth = 1, pt = 3;
void generate_ans(int mx) {
  vector<int> p;
  p.push_back(1);
  for (int i = 0; i < mx; i++) {
    int idx = max(0, int(p.size()) - 2);
    for (int j = idx; j < p.size(); j++)
      grid[pt][p[j]] = grid[p[j]][pt] = grid[pt + 1][p[j]] =
          grid[p[j]][pt + 1] = 1;
    p.push_back(pt);
    p.push_back(pt + 1);
    if (i + 1 == depth)
      grid[pt][2] = grid[2][pt] = grid[pt + 1][2] = grid[2][pt + 1] = 1;
    pt += 2;
  }
  mx = depth - mx;
  if (mx) {
    int idx = max(0, int(p.size()) - 2);
    while (idx < p.size()) {
      int a = p[idx];
      grid[a][37 - mx] = grid[37 - mx][a] = 1;
      idx++;
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> k;
  if (k == 1) {
    cout << "2\nNY\nYN";
    return 0;
  }
  for (int i = 3; i < 36; i++) grid[i][i + 1] = grid[i + 1][i] = 1;
  grid[36][2] = grid[2][36] = 1;
  pt = 38;
  for (int i = 30; i >= 0; i--) {
    if (k & (1 << i)) depth = max(depth, i), generate_ans(i);
  }
  cout << pt - 1 << endl;
  char z[5] = "NY";
  for (int i = 1; i < pt; i++) {
    for (int j = 1; j < pt; j++) printf("%c", z[grid[i][j]]);
    printf("\n");
  }
  return 0;
}

```