Fox and Minimal path

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 G _ij 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: G _ii = ‘N’ and G _ij = G _ji 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.

Solution:

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>

using namespace std;

char g[2010][2010];

int main() {
int k;
scanf("%d", &k);
if (k == 1) {
puts("2");
puts("NY");
puts("YN");
return 0;
}
int mx = 0;
for (int j = 0; j < 30; j++)
    if (k & (1 << j)) {
      mx = j;
    }
  memset(g, 'N', sizeof(g));
  int n = 2;
  int u = 1;
  for (int j = 1; j <= mx; j++) {
    n++;
    g[u][n] = g[n][u] = 'Y';
    int x = n;
    n++;
    g[u][n] = g[n][u] = 'Y';
    int y = n;
    int z;
    if (j == mx) {
      z = 2;
    } else {
      z = ++n;
    }
    g[x][z] = g[z][x] = 'Y';
    g[y][z] = g[z][y] = 'Y';
    u = z;
  }
  for (int j = mx - 1; j >= 0; j--) {
if (k & (1 << j)) {
      int v = 3 * j + 2;
      if (j == 0) v--;
      int len = 2 * (mx - j);
      for (int it = 0; it < len - 1; it++) {
        n++;
        g[v][n] = g[n][v] = 'Y';
        v = n;
      }
      g[v][2] = g[2][v] = 'Y';
    }
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) putchar(g[i][j]);
    putchar('\n');
  }
  return 0;
}