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 a[100][100];
vector<int> ans;
vector<pair<int, int> > edge;
void build(int n) {
  int i, j, u, v;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) a[i][j] = 0;
  for (i = 0; i < ((int)(edge).size()); i++) {
    u = edge[i].first;
    v = edge[i].second;
    a[u][v] = a[v][u] = 1;
  }
}
void pri(int n) {
  int i, j;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (a[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
vector<int> getbit(int n) {
  int pos = 0;
  (ans).clear();
  while (n) {
    if (n & 1) ans.push_back(pos);
    n >>= 1;
    pos++;
  }
  return ans;
}
void addedge(int u, int v) { edge.push_back(pair<int, int>(u, v)); }
int main() {
  int i, m, n, bit, uu, vv, u, v, pos;
  while (scanf("%d", &m) != EOF) {
    if (m == 1) {
      printf("2\n");
      printf("NY\n");
      printf("YN\n");
      continue;
    }
    (edge).clear();
    vector<int> ans = getbit(m);
    sort((ans).begin(), (ans).end());
    bit = ans[((int)(ans).size()) - 1];
    n = bit * 3 + 2;
    for (i = 2; i <= bit; i++) {
      uu = (i - 1) * 3;
      vv = uu + 1;
      u = i * 3;
      v = u + 1;
      addedge(uu, u);
      addedge(uu, v);
      addedge(vv, u);
      addedge(vv, v);
    }
    addedge(1, 3);
    addedge(1, 4);
    addedge(bit * 3, 2);
    addedge(bit * 3 + 1, 2);
    for (i = 1; i < bit; i++) addedge(i * 3 + 2, (i + 1) * 3 + 2);
    addedge(bit * 3 + 2, 2);
    for (i = 0; i < ((int)(ans).size()) - 1; i++) {
      pos = ans[i];
      if (pos == 0)
        addedge(1, 5);
      else {
        uu = pos * 3;
        vv = uu + 1;
        u = (pos + 1) * 3 + 2;
        addedge(uu, u);
        addedge(vv, u);
      }
    }
    printf("%d\n", n);
    build(n);
    pri(n);
  }
  return 0;
}