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>
using namespace std;
inline int init() {
  int now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
inline long long llinit() {
  long long now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int n, c;
int mat[1005][1005];
int len[1005], maxlen;
int depth[1005];
int main() {
  int tmd = 0;
  long long k;
  n = 2;
  k = llinit();
  long long t, tmp = 1, dep = 0;
  t = k;
  maxlen = 0;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 1; j <= 1000; j++) {
      mat[i][j] = 0;
    }
  }
  t--;
  ++c;
  len[c] = 0;
  for (; t; t >>= 1) {
    if (t & 1) {
      ++c;
      len[c] = tmd;
    }
    ++tmd;
    maxlen = max(maxlen, tmd);
    tmp *= 2;
  }
  int lastnode1, lastnode2;
  for (int i = 1; i <= c; i++) {
    dep = 0;
    if (len[i] >= 1) {
      dep = 1;
      lastnode1 = ++n;
      lastnode2 = ++n;
      mat[1][lastnode1] = 1;
      mat[1][lastnode2] = 1;
      for (int j = 2; j <= len[i]; j++) {
        ++n;
        mat[lastnode1][n] = 1;
        mat[lastnode2][n] = 1;
        ++n;
        mat[lastnode1][n] = 1;
        mat[lastnode2][n] = 1;
        lastnode1 = n - 1;
        lastnode2 = n;
      }
      mat[lastnode1][depth[len[i] + 1] + 1] = 1;
      mat[lastnode2][depth[len[i] + 1] + 1] = 1;
    } else {
      ++dep;
      depth[1] = 1;
      ++n;
      mat[1][n] = 1;
      depth[++dep] = n;
      lastnode1 = n;
      for (int j = 2; j <= maxlen; j++) {
        ++n;
        mat[lastnode1][n] = 1;
        depth[++dep] = n;
        lastnode1 = n;
      }
      mat[lastnode1][2] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (mat[i][j] == 1) {
        mat[j][i] = 1;
      }
    }
  }
  printf("%d\n", n);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (mat[i][j] == 1) {
        putchar('Y');
      } else
        putchar('N');
    }
    putchar('\n');
  }
  return 0;
}