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;
char a[5000][5000];
pair<pair<int, int>, vector<vector<int> > > mul(
    pair<pair<int, int>, vector<vector<int> > > g1, int k) {
  int no = g1.first.first;
  int nn = g1.first.first + k + 1;
  vector<vector<int> > m(nn);
  for (int i = 0; i < nn; i++) {
    for (int j = 0; j < nn; j++) {
      if (i < no && j < no) {
        m[i].push_back(g1.second[i][j]);
      } else {
        m[i].push_back(0);
      }
    }
  }
  for (int i = 0; i < k; i++) {
    m[no - 1][no + i] = m[no + i][no - 1] = 1;
    m[nn - 1][no + i] = m[no + i][nn - 1] = 1;
  }
  return make_pair(make_pair(nn, g1.first.second + 2), m);
}
pair<pair<int, int>, vector<vector<int> > > add(
    pair<pair<int, int>, vector<vector<int> > > g1, int k) {
  if (k == 0) {
    return g1;
  }
  g1 = add(g1, k - 1);
  int no = g1.first.first;
  int nn = g1.first.first + g1.first.second - 1;
  vector<vector<int> > m(nn);
  for (int i = 0; i < nn; i++) {
    for (int j = 0; j < nn; j++) {
      if (i < no && j < no) {
        m[i].push_back(g1.second[i][j]);
      } else {
        m[i].push_back(0);
      }
    }
  }
  for (int i = 0; i < g1.first.second; i++) {
    int a = no + i - 1;
    int b = no + i;
    if (i == 0) {
      a = 0;
    }
    if (i == g1.first.second - 1) {
      b = no - 1;
    }
    m[a][b] = m[b][a] = 1;
  }
  for (int i = 0; i < nn; i++)
    if (i != no - 1 && i != nn - 1) {
      swap(m[no - 1][i], m[nn - 1][i]);
      swap(m[i][no - 1], m[i][nn - 1]);
    }
  return make_pair(make_pair(nn, g1.first.second), m);
}
pair<pair<int, int>, vector<vector<int> > > gen(int k) {
  if (k == 1) {
    vector<vector<int> > vv(3);
    vv[0].push_back(0);
    vv[0].push_back(1);
    vv[0].push_back(0);
    vv[1].push_back(1);
    vv[1].push_back(0);
    vv[1].push_back(1);
    vv[2].push_back(0);
    vv[2].push_back(1);
    vv[2].push_back(0);
    return make_pair(make_pair(3, 2), vv);
  }
  pair<pair<int, int>, vector<vector<int> > > g1 = gen(k / 2);
  g1 = mul(g1, 2);
  if (k % 2 == 1) {
    return add(g1, 1);
  } else {
    return g1;
  }
}
int main() {
  int k;
  cin >> k;
  pair<pair<int, int>, vector<vector<int> > > g = gen(k);
  int nn = g.first.first;
  int no = 1;
  for (int i = 0; i < nn; i++)
    if (i != no && i != nn - 1) {
      swap(g.second[no][i], g.second[nn - 1][i]);
      swap(g.second[i][no], g.second[i][nn - 1]);
    }
  printf("%d\n", nn);
  for (int i = 0; i < nn; i++) {
    for (int j = 0; j < nn; j++) {
      if (g.second[i][j] == 0) {
        printf("N");
      } else {
        printf("Y");
      }
    }
    printf("\n");
  }
  return 0;
}