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.
#include <bits/stdc++.h>
using namespace std;
bool g[1010][1010];
int n, k;
vector<int> v1, v2;
void init(void) {
  scanf("%d", &k);
  n = 2;
  v1.push_back(1);
  v2.push_back(2);
}
void addedge(int u, int v) {
  g[u][v] = true;
  g[v][u] = true;
}
void createnode(int x, vector<int> &v) {
  v.clear();
  for (int i = 0; i < (x); i = i + 1) {
    n++;
    assert(n <= 1000);
    v.push_back(n);
  }
}
void filledge(const vector<int> &a, const vector<int> &b) {
  for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
    for (__typeof((b).begin()) jt = (b).begin(); jt != (b).end(); jt++)
      addedge(*it, *jt);
}
void addlayer(int a, int b) {
  vector<vector<int> > v;
  v.clear();
  v.push_back(vector<int>());
  createnode(a, v[0]);
  for (int i = 0; i < (9); i = i + 1) {
    v.push_back(vector<int>());
    if (i < b)
      createnode(10, v[v.size() - 1]);
    else
      createnode(1, v[v.size() - 1]);
  }
  for (int i = 0; i < (v.size() - 1); i = i + 1) filledge(v[i], v[i + 1]);
  filledge(v1, v[0]);
  filledge(v2, v[v.size() - 1]);
}
void process(void) {
  int t = 0;
  while (k > 0) {
    int x = k % 10;
    k = k / 10;
    addlayer(x, t);
    t++;
  }
  printf("%d\n", n);
  for (int i = (1); i <= (n); i = i + 1) {
    for (int j = (1); j <= (n); j = j + 1) {
      if (g[i][j])
        printf("Y");
      else
        printf("N");
    }
    printf("\n");
  }
}
int main(void) {
  init();
  process();
  return 0;
}
