#include <bits/stdc++.h>
const int MAXN = (int)1e6 + 123;
const double eps = 1e-6;
const long long INF = 1000000000000000000ll;
using namespace std;
int k, ans[2000][2000], n, now;
vector<int> b;
pair<int, int> lv[40];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> k;
  if (k == 1) {
    cout << "2\nNY\nYN";
    return 0;
  }
  for (int i = 30; i >= 0; i--)
    if ((k & (1 << i)) > 0) b.push_back(i);
  n = 2;
  lv[0].first = lv[0].second = 1;
  for (int i = 1; i <= b[0]; i++) {
    lv[i].first = ++n;
    lv[i].second = ++n;
  }
  lv[b[0] + 1].first = lv[b[0] + 1].second = 2;
  for (int i = 0; i <= b[0]; i++) {
    ans[lv[i].first][lv[i + 1].first] = ans[lv[i + 1].first][lv[i].first] = 1;
    ans[lv[i].first][lv[i + 1].second] = ans[lv[i + 1].second][lv[i].first] = 1;
    ans[lv[i].second][lv[i + 1].first] = ans[lv[i + 1].first][lv[i].second] = 1;
    ans[lv[i].second][lv[i + 1].second] = ans[lv[i + 1].second][lv[i].second] =
        1;
  }
  for (int i = 1; i < b.size(); i++) {
    now = 1;
    for (int j = 1; j <= b[0] - b[i]; j++) {
      n++;
      ans[now][n] = ans[n][now] = 1;
      now = n;
    }
    int id = b[0] - b[i] + 1;
    ans[now][lv[id].first] = ans[lv[id].first][now] = 1;
    ans[now][lv[id].second] = ans[lv[id].second][now] = 1;
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (ans[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  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>
const int MAXN = (int)1e6 + 123;
const double eps = 1e-6;
const long long INF = 1000000000000000000ll;
using namespace std;
int k, ans[2000][2000], n, now;
vector<int> b;
pair<int, int> lv[40];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> k;
  if (k == 1) {
    cout << "2\nNY\nYN";
    return 0;
  }
  for (int i = 30; i >= 0; i--)
    if ((k & (1 << i)) > 0) b.push_back(i);
  n = 2;
  lv[0].first = lv[0].second = 1;
  for (int i = 1; i <= b[0]; i++) {
    lv[i].first = ++n;
    lv[i].second = ++n;
  }
  lv[b[0] + 1].first = lv[b[0] + 1].second = 2;
  for (int i = 0; i <= b[0]; i++) {
    ans[lv[i].first][lv[i + 1].first] = ans[lv[i + 1].first][lv[i].first] = 1;
    ans[lv[i].first][lv[i + 1].second] = ans[lv[i + 1].second][lv[i].first] = 1;
    ans[lv[i].second][lv[i + 1].first] = ans[lv[i + 1].first][lv[i].second] = 1;
    ans[lv[i].second][lv[i + 1].second] = ans[lv[i + 1].second][lv[i].second] =
        1;
  }
  for (int i = 1; i < b.size(); i++) {
    now = 1;
    for (int j = 1; j <= b[0] - b[i]; j++) {
      n++;
      ans[now][n] = ans[n][now] = 1;
      now = n;
    }
    int id = b[0] - b[i] + 1;
    ans[now][lv[id].first] = ans[lv[id].first][now] = 1;
    ans[now][lv[id].second] = ans[lv[id].second][now] = 1;
  }
  cout << n << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (ans[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}

```