#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9, maxn = (int)1e5 + 1;
const double eps = (double)1e-8;
const int mod = (int)1000000009;
vector<int> a[10001];
int k, x, y, st[1000], xx, step;
bool b[1001][1001];
int main() {
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  x = 1;
  while (k >= (1 << x)) x++;
  x--;
  y = 2;
  a[1].push_back(3);
  for (int i = 2; i <= 2 * x - 1; i++) {
    y++;
    a[y].push_back(y + 1);
  }
  y++;
  a[y].push_back(2);
  y++;
  a[1].push_back(y);
  a[y].push_back(a[3][0]);
  for (int i = 1; i <= x - 1; i++) {
    y++;
    a[(i + 1) * 2].push_back(y);
    a[y].push_back(a[(i + 1) * 2 + 1][0]);
  }
  memset((st), 0, sizeof(st));
  ;
  for (int i = 1; i <= 2 * x - 2; i++) {
    y++;
    a[y].push_back(y + 1);
    if (i & 1) st[i / 2] = y;
  }
  y++;
  a[y].push_back(2);
  st[x - 1] = y;
  xx = x;
  x = 0;
  a[0].push_back(1);
  for (;;) {
    x = a[x][0];
    if (x == 2) break;
    if (x != 1 && (x % 2 > 0))
      x = x;
    else {
      if (x == 1)
        step = 0;
      else
        step = (x / 2 - 1);
      if ((k & (1 << step)) > 0) {
        a[x].push_back(st[step]);
        continue;
      }
    }
  }
  cout << y << endl;
  memset((b), 0, sizeof(b));
  ;
  for (int i = 1; i <= y; i++) {
    for (int j = 0; j < a[i].size(); j++) {
      b[i][a[i][j]] = 1;
      b[a[i][j]][i] = 1;
    }
  }
  for (int i = 1; i <= y; i++) {
    for (int j = 1; j <= y; j++) {
      if (b[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    printf("\n");
  }
  return 0;
}

### 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;
const int inf = (int)1e9, maxn = (int)1e5 + 1;
const double eps = (double)1e-8;
const int mod = (int)1000000009;
vector<int> a[10001];
int k, x, y, st[1000], xx, step;
bool b[1001][1001];
int main() {
  cin >> k;
  if (k == 1) {
    cout << 2 << endl;
    cout << "NY" << endl;
    cout << "YN" << endl;
    return 0;
  }
  x = 1;
  while (k >= (1 << x)) x++;
  x--;
  y = 2;
  a[1].push_back(3);
  for (int i = 2; i <= 2 * x - 1; i++) {
    y++;
    a[y].push_back(y + 1);
  }
  y++;
  a[y].push_back(2);
  y++;
  a[1].push_back(y);
  a[y].push_back(a[3][0]);
  for (int i = 1; i <= x - 1; i++) {
    y++;
    a[(i + 1) * 2].push_back(y);
    a[y].push_back(a[(i + 1) * 2 + 1][0]);
  }
  memset((st), 0, sizeof(st));
  ;
  for (int i = 1; i <= 2 * x - 2; i++) {
    y++;
    a[y].push_back(y + 1);
    if (i & 1) st[i / 2] = y;
  }
  y++;
  a[y].push_back(2);
  st[x - 1] = y;
  xx = x;
  x = 0;
  a[0].push_back(1);
  for (;;) {
    x = a[x][0];
    if (x == 2) break;
    if (x != 1 && (x % 2 > 0))
      x = x;
    else {
      if (x == 1)
        step = 0;
      else
        step = (x / 2 - 1);
      if ((k & (1 << step)) > 0) {
        a[x].push_back(st[step]);
        continue;
      }
    }
  }
  cout << y << endl;
  memset((b), 0, sizeof(b));
  ;
  for (int i = 1; i <= y; i++) {
    for (int j = 0; j < a[i].size(); j++) {
      b[i][a[i][j]] = 1;
      b[a[i][j]][i] = 1;
    }
  }
  for (int i = 1; i <= y; i++) {
    for (int j = 1; j <= y; j++) {
      if (b[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    printf("\n");
  }
  return 0;
}

```