#include <bits/stdc++.h>
using namespace std;
int usedV;
int lastV;
int lastVDist;
int K;
char mp[1007][1007];
void rec(int waysHere, int connTo, int dist) {
  if (waysHere * 2 > K) {
    mp[connTo][2] = 'Y';
    K -= waysHere;
    lastV = connTo;
    lastVDist = dist;
    return;
  }
  int a = ++usedV;
  int b = ++usedV;
  int c = ++usedV;
  mp[connTo][a] = 'Y';
  mp[connTo][b] = 'Y';
  mp[a][c] = 'Y';
  mp[b][c] = 'Y';
  rec(waysHere * 2, c, dist + 2);
  if (waysHere <= K) {
    int last = connTo;
    for (int i = (0); i < (lastVDist - dist - 1); i++) {
      mp[last][++usedV] = 'Y';
      last = usedV;
    }
    mp[last][lastV] = 'Y';
    K -= waysHere;
  }
}
int main() {
  scanf("%d ", &K);
  ;
  for (int i = (0); i < (1007); i++)
    for (int j = (0); j < (1007); j++) mp[i][j] = 'N';
  mp[1][3] = 'Y';
  usedV = 3;
  rec(1, 3, 0);
  for (int i = (0); i < (1007); i++)
    for (int j = (0); j < (1007); j++) {
      if (mp[i][j] == 'Y' || mp[j][i] == 'Y') mp[i][j] = mp[j][i] = 'Y';
    }
  printf("%d\n", usedV);
  for (int i = (0); i < (usedV); i++) {
    for (int j = (0); j < (usedV); j++) {
      printf("%c", mp[i + 1][j + 1]);
    }
    printf("\n");
  }
  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;
int usedV;
int lastV;
int lastVDist;
int K;
char mp[1007][1007];
void rec(int waysHere, int connTo, int dist) {
  if (waysHere * 2 > K) {
    mp[connTo][2] = 'Y';
    K -= waysHere;
    lastV = connTo;
    lastVDist = dist;
    return;
  }
  int a = ++usedV;
  int b = ++usedV;
  int c = ++usedV;
  mp[connTo][a] = 'Y';
  mp[connTo][b] = 'Y';
  mp[a][c] = 'Y';
  mp[b][c] = 'Y';
  rec(waysHere * 2, c, dist + 2);
  if (waysHere <= K) {
    int last = connTo;
    for (int i = (0); i < (lastVDist - dist - 1); i++) {
      mp[last][++usedV] = 'Y';
      last = usedV;
    }
    mp[last][lastV] = 'Y';
    K -= waysHere;
  }
}
int main() {
  scanf("%d ", &K);
  ;
  for (int i = (0); i < (1007); i++)
    for (int j = (0); j < (1007); j++) mp[i][j] = 'N';
  mp[1][3] = 'Y';
  usedV = 3;
  rec(1, 3, 0);
  for (int i = (0); i < (1007); i++)
    for (int j = (0); j < (1007); j++) {
      if (mp[i][j] == 'Y' || mp[j][i] == 'Y') mp[i][j] = mp[j][i] = 'Y';
    }
  printf("%d\n", usedV);
  for (int i = (0); i < (usedV); i++) {
    for (int j = (0); j < (usedV); j++) {
      printf("%c", mp[i + 1][j + 1]);
    }
    printf("\n");
  }
  return 0;
}

```