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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T, class X>
inline bool getbit(T a, X i) {
  T t = 1;
  return ((a & (t << i)) > 0);
}
template <class T, class X>
inline T setbit(T a, X i) {
  T t = 1;
  return (a | (t << i));
}
template <class T, class X>
inline T resetbit(T a, X i) {
  T t = 1;
  return (a & (~(t << i)));
}
inline long long power(long long a, long long b) {
  long long multiply = 1;
  for (int i = (0); i < (b); i++) {
    multiply *= a;
  }
  return multiply;
}
bool graph[1002][1002];
void build(int i, int j) { graph[i][j] = graph[j][i] = true; }
int main() {
  int k;
  cin >> k;
  int node = 3, now = 1;
  int layer = 0;
  vector<pair<int, int> > left;
  while (k > 1) {
    if (k % 2) {
      left.push_back(make_pair(now, layer));
      k--;
    }
    build(now, node);
    build(now, node + 1);
    now = node + 2;
    build(now, node);
    build(now, node + 1);
    node = now + 1;
    k /= 2;
    layer += 2;
  }
  build(2, now);
  for (int j = (0); j < (left.size()); j++) {
    int now = left[j].first, where = left[j].second;
    while (where < layer) {
      build(now, node);
      where++;
      now = node++;
    }
    build(now, 2);
  }
  cout << node - 1 << endl;
  for (int i = (1); i < (node); i++) {
    for (int j = (1); j < (node); j++) {
      if (graph[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
  return 0;
}