#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 10;
const int inf = 1000000007;
const int mod = 1000000007;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
vector<int> e[maxn];
int d[maxn];
char c[maxn][maxn];
int main() {
  int k;
  while (cin >> k) {
    for (int i = 0; i < maxn; i++) e[i].clear();
    int cur = 1, nw = 3;
    while (k > 1) {
      if (k % 2 == 0) {
        int a1 = nw++;
        int a2 = nw++;
        int a3 = nw++;
        d[a1] = d[cur] + 1;
        d[a2] = d[cur] + 1;
        d[a3] = d[cur] + 2;
        e[cur].push_back(a1);
        e[cur].push_back(a2);
        e[a1].push_back(a3);
        e[a2].push_back(a3);
        cur = a3;
        k /= 2;
      } else {
        int a1 = nw++;
        e[cur].push_back(a1);
        d[a1] = d[cur] + 1;
        k--;
      }
    }
    e[cur].push_back(2);
    d[2] = d[cur] + 1;
    for (int i = 1; i < nw; i++) {
      if (e[i].empty() && i != 2) {
        if (d[i] + 1 < d[2]) {
          int a1 = nw++;
          d[a1] = d[i] + 1;
          e[i].push_back(a1);
        } else
          e[i].push_back(2);
      }
    }
    for (int i = 1; i < nw; i++)
      for (int j = 1; j < nw; j++) c[i][j] = 'N';
    for (int i = 1; i < nw; i++)
      for (int j = 0; j < e[i].size(); j++) c[i][e[i][j]] = c[e[i][j]][i] = 'Y';
    cout << nw - 1 << endl;
    for (int i = 1; i < nw; i++) {
      for (int j = 1; j < nw; j++) {
        cout << c[i][j];
      }
      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>
using namespace std;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 10;
const int inf = 1000000007;
const int mod = 1000000007;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
vector<int> e[maxn];
int d[maxn];
char c[maxn][maxn];
int main() {
  int k;
  while (cin >> k) {
    for (int i = 0; i < maxn; i++) e[i].clear();
    int cur = 1, nw = 3;
    while (k > 1) {
      if (k % 2 == 0) {
        int a1 = nw++;
        int a2 = nw++;
        int a3 = nw++;
        d[a1] = d[cur] + 1;
        d[a2] = d[cur] + 1;
        d[a3] = d[cur] + 2;
        e[cur].push_back(a1);
        e[cur].push_back(a2);
        e[a1].push_back(a3);
        e[a2].push_back(a3);
        cur = a3;
        k /= 2;
      } else {
        int a1 = nw++;
        e[cur].push_back(a1);
        d[a1] = d[cur] + 1;
        k--;
      }
    }
    e[cur].push_back(2);
    d[2] = d[cur] + 1;
    for (int i = 1; i < nw; i++) {
      if (e[i].empty() && i != 2) {
        if (d[i] + 1 < d[2]) {
          int a1 = nw++;
          d[a1] = d[i] + 1;
          e[i].push_back(a1);
        } else
          e[i].push_back(2);
      }
    }
    for (int i = 1; i < nw; i++)
      for (int j = 1; j < nw; j++) c[i][j] = 'N';
    for (int i = 1; i < nw; i++)
      for (int j = 0; j < e[i].size(); j++) c[i][e[i][j]] = c[e[i][j]][i] = 'Y';
    cout << nw - 1 << endl;
    for (int i = 1; i < nw; i++) {
      for (int j = 1; j < nw; j++) {
        cout << c[i][j];
      }
      cout << endl;
    }
  }
  return 0;
}

```