#include <bits/stdc++.h>
using namespace std;
int k;
int mypow[10];
int mymax = 0;
char ok = 0;
int mycode(int i, int level) {
  int z = level * 10 + i + 2;
  if (ok) mymax = max(z, mymax);
  return z;
}
char mat[1050][1050];
void join(int k, int level, int node) {
  for (int i = 0; i < k; i++) {
    mat[mycode(i, level)][node] = 'Y';
    mat[node][mycode(i, level)] = 'Y';
  }
}
int CON = 200;
int MAXLEVEL = 9;
void disperse(int k) {
  int level = 0;
  while (k > 0) {
    if (k < 10)
      join(k % 10, level, 0);
    else
      join(k % 10, level, level + CON);
    mat[level + CON + 1][level + CON] = 'Y';
    mat[level + CON][level + CON + 1] = 'Y';
    k /= 10;
    level++;
  }
  level -= 2;
  mat[level + CON][0] = 'Y';
  mat[0][level + CON] = 'Y';
}
int main() {
  int x;
  scanf("%d", &x);
  mypow[0] = 1;
  for (int i = 1; i < 10; i++) mypow[i] = mypow[i - 1] * 10;
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= 1000; j++) mat[i][j] = 'N';
  for (int level = 0; level < MAXLEVEL; level++)
    for (int i = 0; i < 10; i++) {
      join(10, level + 1, mycode(i, level));
    }
  ok = 1;
  disperse(x);
  join(10, 0, 1);
  mymax++;
  mymax = 10 + CON;
  printf("%d\n", mymax);
  for (int i = 0; i < mymax; i++) {
    for (int j = 0; j < mymax; j++) printf("%c", mat[i][j]);
    printf("\n");
  }
}

### Prompt

Create a solution in cpp for 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;
int k;
int mypow[10];
int mymax = 0;
char ok = 0;
int mycode(int i, int level) {
  int z = level * 10 + i + 2;
  if (ok) mymax = max(z, mymax);
  return z;
}
char mat[1050][1050];
void join(int k, int level, int node) {
  for (int i = 0; i < k; i++) {
    mat[mycode(i, level)][node] = 'Y';
    mat[node][mycode(i, level)] = 'Y';
  }
}
int CON = 200;
int MAXLEVEL = 9;
void disperse(int k) {
  int level = 0;
  while (k > 0) {
    if (k < 10)
      join(k % 10, level, 0);
    else
      join(k % 10, level, level + CON);
    mat[level + CON + 1][level + CON] = 'Y';
    mat[level + CON][level + CON + 1] = 'Y';
    k /= 10;
    level++;
  }
  level -= 2;
  mat[level + CON][0] = 'Y';
  mat[0][level + CON] = 'Y';
}
int main() {
  int x;
  scanf("%d", &x);
  mypow[0] = 1;
  for (int i = 1; i < 10; i++) mypow[i] = mypow[i - 1] * 10;
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= 1000; j++) mat[i][j] = 'N';
  for (int level = 0; level < MAXLEVEL; level++)
    for (int i = 0; i < 10; i++) {
      join(10, level + 1, mycode(i, level));
    }
  ok = 1;
  disperse(x);
  join(10, 0, 1);
  mymax++;
  mymax = 10 + CON;
  printf("%d\n", mymax);
  for (int i = 0; i < mymax; i++) {
    for (int j = 0; j < mymax; j++) printf("%c", mat[i][j]);
    printf("\n");
  }
}

```