#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1011;
int e[SIZE][SIZE], id;
void add(int x, int y) { e[x][y] = e[y][x] = 1; }
int go(int st, int ed, int num) {
  if (num == 1) {
    add(st, ed);
    return 1;
  }
  if (num == 2) {
    add(st, id);
    add(id++, ed);
    add(st, id);
    add(id++, ed);
    return 2;
  }
  if (num & 1) {
    int len = go(st, ed, num - 1);
    add(st, id);
    for (int i = 1; i < len; i++) {
      if (i != len - 1) {
        add(id, id + 1);
        id++;
      } else {
        add(id, ed);
        id++;
      }
    }
    return len;
  } else {
    int mid = id++;
    int len = go(st, mid, num / 2);
    len += go(mid, ed, 2);
    return len;
  }
}
int main() {
  int(K);
  scanf("%d", &K);
  id = 3;
  go(1, 2, K);
  printf("%d\n", id - 1);
  for (int i = (1); i < (id); ++i) {
    for (int j = (1); j < (id); ++j) {
      if (e[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
  return 0;
}

### Prompt

Please formulate 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 SIZE = 1011;
int e[SIZE][SIZE], id;
void add(int x, int y) { e[x][y] = e[y][x] = 1; }
int go(int st, int ed, int num) {
  if (num == 1) {
    add(st, ed);
    return 1;
  }
  if (num == 2) {
    add(st, id);
    add(id++, ed);
    add(st, id);
    add(id++, ed);
    return 2;
  }
  if (num & 1) {
    int len = go(st, ed, num - 1);
    add(st, id);
    for (int i = 1; i < len; i++) {
      if (i != len - 1) {
        add(id, id + 1);
        id++;
      } else {
        add(id, ed);
        id++;
      }
    }
    return len;
  } else {
    int mid = id++;
    int len = go(st, mid, num / 2);
    len += go(mid, ed, 2);
    return len;
  }
}
int main() {
  int(K);
  scanf("%d", &K);
  id = 3;
  go(1, 2, K);
  printf("%d\n", id - 1);
  for (int i = (1); i < (id); ++i) {
    for (int j = (1); j < (id); ++j) {
      if (e[i][j])
        putchar('Y');
      else
        putchar('N');
    }
    puts("");
  }
  return 0;
}

```