#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
bool f[1001][1001];
vector<int> st;
int now;
vector<int> mt[55];
void add(int x) {
  int i, j;
  for (i = now; i < now + x; i++) {
    for (j = 0; j < st.size(); j++) {
      f[st[j]][i] = f[i][st[j]] = 1;
    }
  }
  st.clear();
  for (i = now; i < now + x; i++) {
    st.push_back(i);
  }
  now += x;
}
void ad(int x) {
  int i;
  for (i = 0; i < st.size(); i++) {
    f[now][st[i]] = f[st[i]][now] = 1;
  }
  for (i = now + 1; i < now + x; i++) {
    f[i][i - 1] = f[i - 1][i] = 1;
  }
  now += x;
  f[now - 1][2] = f[2][now - 1] = 1;
}
int a[55];
int len;
int main() {
  int k;
  int end = sqrt(1e9) + 1;
  while (cin >> k) {
    memset(f, 0, sizeof(f));
    int i, j;
    int n = 2;
    len = 0;
    while (k) {
      len++;
      if (k & 1) {
        a[len] = 1;
      } else
        a[len] = 0;
      k >>= 1;
    }
    now = 3;
    st.push_back(1);
    for (i = 1; i < len; i++) {
      if (a[i]) {
        ad(len - i);
      }
      add(2);
    }
    for (i = 0; i < st.size(); i++) {
      f[st[i]][2] = f[2][st[i]] = 1;
    }
    n = now - 1;
    cout << n << endl;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if (f[i][j])
          putchar('Y');
        else
          putchar('N');
      }
      putchar('\n');
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 double pi = acos(-1.0);
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
bool f[1001][1001];
vector<int> st;
int now;
vector<int> mt[55];
void add(int x) {
  int i, j;
  for (i = now; i < now + x; i++) {
    for (j = 0; j < st.size(); j++) {
      f[st[j]][i] = f[i][st[j]] = 1;
    }
  }
  st.clear();
  for (i = now; i < now + x; i++) {
    st.push_back(i);
  }
  now += x;
}
void ad(int x) {
  int i;
  for (i = 0; i < st.size(); i++) {
    f[now][st[i]] = f[st[i]][now] = 1;
  }
  for (i = now + 1; i < now + x; i++) {
    f[i][i - 1] = f[i - 1][i] = 1;
  }
  now += x;
  f[now - 1][2] = f[2][now - 1] = 1;
}
int a[55];
int len;
int main() {
  int k;
  int end = sqrt(1e9) + 1;
  while (cin >> k) {
    memset(f, 0, sizeof(f));
    int i, j;
    int n = 2;
    len = 0;
    while (k) {
      len++;
      if (k & 1) {
        a[len] = 1;
      } else
        a[len] = 0;
      k >>= 1;
    }
    now = 3;
    st.push_back(1);
    for (i = 1; i < len; i++) {
      if (a[i]) {
        ad(len - i);
      }
      add(2);
    }
    for (i = 0; i < st.size(); i++) {
      f[st[i]][2] = f[2][st[i]] = 1;
    }
    n = now - 1;
    cout << n << endl;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if (f[i][j])
          putchar('Y');
        else
          putchar('N');
      }
      putchar('\n');
    }
  }
  return 0;
}

```