#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
int k;
bool a[N][N];
int n = 0;
int st[N];
int stSZ = 0;
int nextVertex() {
  if (n < N) {
    n++;
    return n - 1;
  }
  exit(-1);
}
int main() {
  scanf("%d", &k);
  int from = nextVertex();
  int to = nextVertex();
  st[stSZ++] = from;
  map<int, int> branches;
  int len = 0;
  while (k > 0) {
    if (k == 1) {
      len++;
      for (int i = 0; i < stSZ; i++) {
        int v = st[i];
        a[to][v] = a[v][to] = true;
      }
      break;
    }
    if (k % 2 == 0) {
      int u = nextVertex(), v = nextVertex();
      for (int i = 0; i < stSZ; i++) {
        a[u][st[i]] = a[st[i]][u] = true;
        a[v][st[i]] = a[st[i]][v] = true;
      }
      stSZ = 0;
      st[stSZ++] = u;
      st[stSZ++] = v;
      len++;
      k /= 2;
    } else {
      int v = nextVertex();
      for (int i = 0; i < stSZ; i++) {
        a[v][st[i]] = a[st[i]][v] = true;
      }
      branches[len + 1] = v;
      k--;
    }
  }
  for (map<int, int>::iterator it = branches.begin(); it != branches.end();
       it++) {
    int i = it->first;
    int v = it->second;
    while (i != len - 1) {
      int u = nextVertex();
      a[u][v] = a[v][u] = true;
      v = u;
      i++;
    }
    a[v][to] = a[to][v] = true;
  }
  printf("%d\n", n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf(a[i][j] ? "Y" : "N");
    }
    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 = 1000;
int k;
bool a[N][N];
int n = 0;
int st[N];
int stSZ = 0;
int nextVertex() {
  if (n < N) {
    n++;
    return n - 1;
  }
  exit(-1);
}
int main() {
  scanf("%d", &k);
  int from = nextVertex();
  int to = nextVertex();
  st[stSZ++] = from;
  map<int, int> branches;
  int len = 0;
  while (k > 0) {
    if (k == 1) {
      len++;
      for (int i = 0; i < stSZ; i++) {
        int v = st[i];
        a[to][v] = a[v][to] = true;
      }
      break;
    }
    if (k % 2 == 0) {
      int u = nextVertex(), v = nextVertex();
      for (int i = 0; i < stSZ; i++) {
        a[u][st[i]] = a[st[i]][u] = true;
        a[v][st[i]] = a[st[i]][v] = true;
      }
      stSZ = 0;
      st[stSZ++] = u;
      st[stSZ++] = v;
      len++;
      k /= 2;
    } else {
      int v = nextVertex();
      for (int i = 0; i < stSZ; i++) {
        a[v][st[i]] = a[st[i]][v] = true;
      }
      branches[len + 1] = v;
      k--;
    }
  }
  for (map<int, int>::iterator it = branches.begin(); it != branches.end();
       it++) {
    int i = it->first;
    int v = it->second;
    while (i != len - 1) {
      int u = nextVertex();
      a[u][v] = a[v][u] = true;
      v = u;
      i++;
    }
    a[v][to] = a[to][v] = true;
  }
  printf("%d\n", n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf(a[i][j] ? "Y" : "N");
    }
    printf("\n");
  }
  return 0;
}

```