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.
#include <bits/stdc++.h>
using namespace std;
struct Node {};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
int k = 0;
int n = 2001 - 1;
char grid[2001][2001];
void connect(int x, int y) {
  grid[x][y] = 'Y';
  grid[y][x] = 'Y';
  return;
}
int main() {
  scanf("%d", &k);
  for (int i = 0; i < n; i++) {
    int j = 0;
    for (j = 0; j < n; j++) {
      grid[i][j] = 'N';
    }
    grid[i][j] = '\0';
  }
  int base = 10;
  vector<int> exp;
  vector<int> extra;
  exp.clear();
  extra.clear();
  while (true) {
    int e = int(log10(double(k)));
    if (e < 2) {
      break;
    }
    exp.push_back(e);
    extra.push_back(k / pow(base, e - 1));
    k -= (int)pow(base, e - 1) * (int)(k / pow(base, e - 1));
  }
  int layer;
  if (exp.size() == 0) {
    layer = 1;
  } else {
    layer = exp[0] * 2 - 1;
  }
  int anchor = 0;
  int pos = 2;
  for (int i = 0; i < exp.size(); i++) {
    anchor = 0;
    int e = exp[i];
    for (int j = 0; j < layer - (e * 2 - 1); j++) {
      connect(anchor, pos);
      anchor = pos;
      pos++;
    }
    for (int j = 1; j <= e; j++) {
      if (j == e) {
        for (int b = 0; b < extra[i]; b++) {
          connect(anchor, pos + b);
        }
        for (int b = 0; b < extra[i]; b++) {
          connect(1, pos + b);
        }
        pos += extra[i];
      } else {
        for (int b = 0; b < base; b++) {
          connect(anchor, pos + b);
        }
        anchor = pos + base;
        for (int b = 0; b < base; b++) {
          connect(anchor, pos + b);
        }
        pos = anchor + 1;
      }
    }
  }
  if (k > 0) {
    anchor = 0;
    for (int i = 0; i < layer - 1; i++) {
      connect(anchor, pos);
      anchor = pos;
      pos += 1;
    }
    for (int j = 0; j < k; j++) {
      connect(anchor, pos);
      connect(1, pos);
      pos++;
    }
  }
  printf("%d\n", pos);
  for (int i = 0; i < pos; i++) {
    for (int j = 0; j < pos; j++) {
      printf("%c", grid[i][j]);
    }
    printf("\n");
  }
  return 0;
}
