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;
bool a[3001][3001], two[3001];
int last[3001];
int main() {
  int k;
  scanf("%d", &k);
  if (k == 1) {
    puts("2");
    puts("NY");
    puts("YN");
    return 0;
  }
  int cnt;
  for (cnt = 0; (1 << cnt) <= k; ++cnt)
    ;
  --cnt;
  int dist = cnt;
  int u = 2;
  while (k) {
    for (cnt = 0; (1 << cnt) <= k; ++cnt)
      ;
    --cnt;
    k -= (1 << cnt);
    bool was = 0;
    if (!cnt) was = 1;
    bool twoVertexes = 0;
    ++u;
    a[1][u] = a[u][1] = 1;
    if (cnt) {
      ++u;
      a[1][u] = a[u][1] = 1;
      twoVertexes = 1;
    }
    int cur = cnt;
    if (!cnt) ++cur;
    --cnt;
    while (cnt > 0) {
      --cnt;
      ++u;
      a[u - 1][u] = a[u][u - 1] = 1;
      a[u - 2][u] = a[u][u - 2] = 1;
      ++u;
      a[u - 2][u] = a[u][u - 2] = 1;
      a[u - 3][u] = a[u][u - 3] = 1;
      twoVertexes = 1;
    }
    if (cur < dist) {
      if (twoVertexes) {
        last[u] = last[u - 1] = dist - cur;
        two[u] = 1;
      } else
        last[u] = dist - cur;
    } else {
      if (twoVertexes) {
        a[u][2] = a[2][u] = 1;
        a[u - 1][2] = a[2][u - 1] = 1;
      } else
        a[u][2] = a[2][u] = 1;
    }
  }
  int numv = u;
  if (last[numv]) {
    int st;
    while (last[numv]) {
      ++u;
      --last[numv];
      a[u][u - 1] = a[u - 1][u] = 1;
    }
    if (two[numv]) {
      st = u - (last[numv - 1]) + 1;
      a[numv - 1][st] = a[st][numv - 1] = 1;
      last[numv - 1] = 0;
    }
    for (int i = 1; i <= numv; i++) {
      if (last[i]) {
        st = u - (last[i]) + 1;
        a[i][st] = a[st][i] = 1;
      }
    }
    a[u][2] = a[2][u] = 1;
  }
  printf("%d\n", u);
  for (int i = 1; i <= u; i++) {
    for (int j = 1; j <= u; j++) printf("%c", (a[i][j]) ? 'Y' : 'N');
    printf("\n");
  }
  return 0;
}