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;
const int MAX_N = 7 + 1000;
bool local = false;
bool tab[MAX_N][MAX_N];
int lowbit(int d) { return d & -d; }
int tot = 2;
int pos[100];
int sou[100];
void link(int a, int b) { tab[a][b] = tab[b][a] = true; }
void init() {
  pos[30] = ++tot;
  for (int i = (0); i < (30); i++) {
    link(tot, tot + 1);
    link(tot + 1, tot + 3);
    link(tot, tot + 2);
    link(tot + 2, tot + 3);
    tot += 3;
    pos[29 - i] = tot;
  }
  link(tot, 2);
  int pre = 1;
  sou[0] = 1;
  for (int i = (1); i < (61); i++) {
    link(pre, ++tot);
    pre = tot;
    sou[i] = tot;
  }
}
void solve(int bit) { link(sou[60 - bit * 2], pos[bit]); }
vector<int> G[MAX_N];
int dis[MAX_N];
int pat[MAX_N];
bool check(int des) {
  for (int i = (1); i < (tot + 1); i++)
    for (int j = (1); j < (tot + 1); j++)
      if (tab[i][j]) G[i].push_back(j);
  memset(dis, -1, sizeof(dis));
  memset(pat, 0, sizeof(pat));
  queue<int> que;
  dis[1] = 0;
  pat[1] = 1;
  for (que.push(1); !que.empty(); que.pop()) {
    int u = que.front();
    for (int i = (0); i < (((int)(G[u]).size())); i++) {
      int v = G[u][i];
      if (!~dis[v]) {
        pat[v] = pat[u];
        dis[v] = dis[u] + 1;
        que.push(v);
      } else if (dis[u] + 1 == dis[v]) {
        pat[v] += pat[u];
      }
    }
  }
  cout << "des"
       << " => " << des << endl;
  cout << "pat[2]"
       << " => " << pat[2] << endl;
  return des == pat[2];
}
void test() {
  cout << "(1 << 29) - 1"
       << " => " << (1 << 29) - 1 << endl;
  exit(0);
}
int main() {
  int des;
  cin >> des;
  init();
  for (int num = des; num; num -= lowbit(num)) {
    int lb = lowbit(num);
    int bit = -1;
    for (int tmp = lb; tmp; tmp >>= 1) bit++;
    solve(bit);
  }
  printf("%d\n", tot);
  if (!local)
    for (int i = (1); i < (tot + 1); i++) {
      for (int j = (1); j < (tot + 1); j++) putchar(tab[i][j] ? 'Y' : 'N');
      printf("\n");
    }
  if (local) puts(check(des) ? "AC" : "WA");
}
