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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
char A[1111][1111];
int K, N;
void WriteOutput() {
  int i, j;
  printf("%d\n", N);
  for (i = 1; i <= N; i++) {
    for (j = 1; j <= N; j++)
      if (A[i][j])
        printf("Y");
      else
        printf("N");
    printf("\n");
  }
}
int l[1111][1111], nlevel[1111];
int need[1111], level[1111];
int q[1111], li, ls, i, lmax;
int GetNode(int lev, int nneed, int excl) {
  int i;
  for (i = 0; i < nlevel[lev]; i++)
    if (l[lev][i] != excl && need[l[lev][i]] == nneed) break;
  if (i < nlevel[lev]) return l[lev][i];
  N++;
  l[lev][nlevel[lev]++] = N;
  level[N] = lev;
  need[N] = nneed;
  ls++;
  q[ls] = N;
  if (lev > lmax) lmax = lev;
  return N;
}
int main() {
  scanf("%d", &K);
  if (K == 1) {
    N = 2;
    A[1][2] = A[2][1] = 1;
  } else {
    N = 2;
    need[2] = K;
    li = 0;
    ls = 0;
    q[0] = 2;
    nlevel[0] = 1;
    l[0][0] = 2;
    level[2] = lmax = 0;
    while (li <= ls) {
      i = q[li];
      li++;
      if (need[i] == 1) continue;
      int x = GetNode(level[i] + 1, need[i] / 2, 0);
      int y = GetNode(level[i] + 1, need[i] - (need[i] / 2), x);
      A[x][i] = A[i][x] = A[y][i] = A[i][y] = 1;
      if (0)
        fprintf(stderr, "i=%d (need=%d) new edges: %d (%d) %d (%d)\n", i,
                need[i], x, need[x], y, need[y]);
    }
    for (i = 1; i <= N; i++)
      if (need[i] == 1 && level[i] != lmax) {
        int x = i;
        need[i] = 0;
        while (level[x] < lmax) {
          N++;
          level[N] = level[x] + 1;
          A[N][x] = A[x][N] = 1;
          if (0) fprintf(stderr, "new edge %d-%d\n", x, N);
          x = N;
          need[x] = 1;
        }
      }
    for (i = 1; i <= N; i++)
      if (need[i] == 1) A[1][i] = A[i][1] = 1;
  }
  WriteOutput();
  return 0;
}