#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
bool mtx[N][N];
int idx = 3;
void create(int x, int y) {
  mtx[x][y] = mtx[y][x] = true;
  return;
}
void solve(int st, int ten, int digit, int cnt) {
  if (cnt == 25) {
    if (digit) {
      while (digit--) {
        int ed = ++idx;
        create(st, ed);
        create(ed, 1);
      }
    } else {
      int ed = ++idx;
      create(st, ed);
      create(ed, 1);
    }
    return;
  }
  if (ten) {
    int ed = ++idx;
    for (int i = 0; i < 10; i++) {
      ++idx;
      create(st, idx);
      create(idx, ed);
    }
    solve(ed, ten - 1, digit, cnt + 2);
  } else {
    int ed = ++idx;
    create(st, ed);
    solve(ed, ten, digit, cnt + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int ten = 0;
  while (n) {
    int d = n % 10;
    if (d) solve(0, ten, d, 0);
    n /= 10;
    ten++;
  }
  idx++;
  printf("%d\n", idx);
  for (int i = 0; i < idx; i++) {
    for (int j = 0; j < idx; j++) {
      putchar(mtx[i][j] ? 'Y' : '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 N = 1000;
bool mtx[N][N];
int idx = 3;
void create(int x, int y) {
  mtx[x][y] = mtx[y][x] = true;
  return;
}
void solve(int st, int ten, int digit, int cnt) {
  if (cnt == 25) {
    if (digit) {
      while (digit--) {
        int ed = ++idx;
        create(st, ed);
        create(ed, 1);
      }
    } else {
      int ed = ++idx;
      create(st, ed);
      create(ed, 1);
    }
    return;
  }
  if (ten) {
    int ed = ++idx;
    for (int i = 0; i < 10; i++) {
      ++idx;
      create(st, idx);
      create(idx, ed);
    }
    solve(ed, ten - 1, digit, cnt + 2);
  } else {
    int ed = ++idx;
    create(st, ed);
    solve(ed, ten, digit, cnt + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int ten = 0;
  while (n) {
    int d = n % 10;
    if (d) solve(0, ten, d, 0);
    n /= 10;
    ten++;
  }
  idx++;
  printf("%d\n", idx);
  for (int i = 0; i < idx; i++) {
    for (int j = 0; j < idx; j++) {
      putchar(mtx[i][j] ? 'Y' : 'N');
    }
    puts("");
  }
  return 0;
}

```