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;
int ma[4000][4000];
int num;
void makeComp(vector<int>& cha) {
  if (cha.empty()) {
    cha.push_back(num);
    num += 4;
  } else {
    cha.push_back(num - 1);
    num += 3;
  }
}
vector<int> makeChain(int order) {
  vector<int> cha;
  for (int i = 0; i < order; ++i) makeComp(cha);
  return cha;
}
void add(int u, int v) { ma[u][v] = ma[v][u] = 1; }
void makeLine(int ini, int which, int largest) {
  int last = ini + (which - 1) * 3;
  for (int i = which; i < largest + 1; ++i) {
    add(last, num);
    add(num, num + 1);
    last = num + 1;
    num += 2;
  }
  add(last, 2);
}
void faz(int k) {
  memset(ma, false, sizeof ma);
  num = 3;
  if (k == 1) return add(1, 2), void();
  vector<int> largest;
  for (int i = 31; i > 0; --i) {
    if ((1 << i) & k) {
      largest = makeChain(i);
      break;
    }
  }
  int last = 1;
  for (int j = 0; j < largest.size(); ++j) {
    int a = largest[j];
    if (last != a) add(last, a);
    add(a, a + 1), add(a, a + 2), add(a + 1, a + 3), add(a + 2, a + 3);
    last = a + 3;
  }
  add(last, 2);
  for (int i = largest.size() - 1; i > 0; --i) {
    if ((1 << i) & k) {
      makeLine(3, i + 1, largest.size());
    }
  }
  if (k & 1) {
    makeLine(3, 1, largest.size());
  }
}
int main() {
  int k;
  cin >> k;
  faz(k);
  cout << num - 1 << endl;
  for (int i = 1; i < num; ++i) {
    for (int j = 1; j < num; ++j) cout << (ma[i][j] ? "Y" : "N");
    cout << endl;
  }
}