#include <bits/stdc++.h>
using namespace std;
int G[1005][1005];
int main() {
  int k;
  scanf("%d", &k);
  int mxBit = 0;
  for (int i = 31; i >= 0; i--) {
    if (k & (1 << i)) {
      mxBit = i;
      break;
    }
  }
  G[0][2] = G[2][0] = true;
  int nodeCount = 3;
  int mxLength = 0;
  vector<int> anchors;
  anchors.push_back(2);
  for (int i = 0; i < mxBit; i += 1) {
    int curNode = anchors.back();
    int n1 = nodeCount++;
    int n2 = nodeCount++;
    int n3 = nodeCount++;
    G[curNode][n1] = G[n1][curNode] = true;
    G[curNode][n2] = G[n2][curNode] = true;
    G[n1][n3] = G[n3][n1] = true;
    G[n2][n3] = G[n3][n2] = true;
    anchors.push_back(n3);
    mxLength += 2;
  }
  {
    int curNode = anchors.back();
    G[curNode][1] = G[1][curNode] = true;
  }
  for (int i = mxBit - 1; i >= 0; i--) {
    if (k & (1 << i)) {
      int curNode = anchors[i];
      int remLength = mxLength - 2 * i;
      for (int j = 0; j < remLength; j += 1) {
        int n1 = nodeCount++;
        G[curNode][n1] = G[n1][curNode] = true;
        curNode = n1;
      }
      G[curNode][1] = G[1][curNode] = true;
    }
  }
  printf("%d\n", nodeCount);
  for (int i = 0; i < nodeCount; i += 1) {
    for (int j = 0; j < nodeCount; j += 1) {
      if (G[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 G[1005][1005];
int main() {
  int k;
  scanf("%d", &k);
  int mxBit = 0;
  for (int i = 31; i >= 0; i--) {
    if (k & (1 << i)) {
      mxBit = i;
      break;
    }
  }
  G[0][2] = G[2][0] = true;
  int nodeCount = 3;
  int mxLength = 0;
  vector<int> anchors;
  anchors.push_back(2);
  for (int i = 0; i < mxBit; i += 1) {
    int curNode = anchors.back();
    int n1 = nodeCount++;
    int n2 = nodeCount++;
    int n3 = nodeCount++;
    G[curNode][n1] = G[n1][curNode] = true;
    G[curNode][n2] = G[n2][curNode] = true;
    G[n1][n3] = G[n3][n1] = true;
    G[n2][n3] = G[n3][n2] = true;
    anchors.push_back(n3);
    mxLength += 2;
  }
  {
    int curNode = anchors.back();
    G[curNode][1] = G[1][curNode] = true;
  }
  for (int i = mxBit - 1; i >= 0; i--) {
    if (k & (1 << i)) {
      int curNode = anchors[i];
      int remLength = mxLength - 2 * i;
      for (int j = 0; j < remLength; j += 1) {
        int n1 = nodeCount++;
        G[curNode][n1] = G[n1][curNode] = true;
        curNode = n1;
      }
      G[curNode][1] = G[1][curNode] = true;
    }
  }
  printf("%d\n", nodeCount);
  for (int i = 0; i < nodeCount; i += 1) {
    for (int j = 0; j < nodeCount; j += 1) {
      if (G[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
  return 0;
}

```