#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e3 + 3;
int K;
int N, L;
char adj[MAXN][MAXN];
void join(int v, int u) { adj[v][u] = adj[u][v] = 'Y'; }
void mkpath(int v, int l, int u) {
  if (l == 1) {
    join(u, v);
  } else {
    join(v, N);
    N++;
    mkpath(N - 1, l - 1, u);
  }
}
int spn[MAXN], spl[MAXN];
int main() {
  scanf("%d", &K);
  memset(adj, 'N', sizeof(adj));
  N = 2;
  if (K == 1) {
    adj[0][1] = adj[1][0] = 'Y';
  } else {
    int maxb = 0;
    for (int i = 0; i < (int)31; i++)
      if (K & (1 << i)) maxb = i;
    L = 2 * maxb;
    int l = 0, v = 1, e = 0;
    spn[e] = v;
    spl[e] = l;
    while (l + 2 < L) {
      join(v, N);
      join(v, N + 1);
      join(N, N + 2);
      join(N + 1, N + 2);
      v = N + 2;
      N += 3;
      l += 2;
      e++;
      spn[e] = v;
      spl[e] = l;
    }
    join(v, N);
    join(v, N + 1);
    join(N, 0);
    join(N + 1, 0);
    N += 2;
    e++;
    spn[e] = v;
    spl[e] = l;
    for (int i = 0; i < (int)maxb; i++)
      if (K & (1 << i)) mkpath(0, L - spl[i], spn[i]);
  }
  for (int i = 0; i < (int)N; i++) {
  }
  printf("%d\n", N);
  for (int i = 0; i < (int)N; i++) {
    for (int j = 0; j < (int)N; j++) printf("%c", adj[i][j]);
    printf("\n");
  }
}

### Prompt

In CPP, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e3 + 3;
int K;
int N, L;
char adj[MAXN][MAXN];
void join(int v, int u) { adj[v][u] = adj[u][v] = 'Y'; }
void mkpath(int v, int l, int u) {
  if (l == 1) {
    join(u, v);
  } else {
    join(v, N);
    N++;
    mkpath(N - 1, l - 1, u);
  }
}
int spn[MAXN], spl[MAXN];
int main() {
  scanf("%d", &K);
  memset(adj, 'N', sizeof(adj));
  N = 2;
  if (K == 1) {
    adj[0][1] = adj[1][0] = 'Y';
  } else {
    int maxb = 0;
    for (int i = 0; i < (int)31; i++)
      if (K & (1 << i)) maxb = i;
    L = 2 * maxb;
    int l = 0, v = 1, e = 0;
    spn[e] = v;
    spl[e] = l;
    while (l + 2 < L) {
      join(v, N);
      join(v, N + 1);
      join(N, N + 2);
      join(N + 1, N + 2);
      v = N + 2;
      N += 3;
      l += 2;
      e++;
      spn[e] = v;
      spl[e] = l;
    }
    join(v, N);
    join(v, N + 1);
    join(N, 0);
    join(N + 1, 0);
    N += 2;
    e++;
    spn[e] = v;
    spl[e] = l;
    for (int i = 0; i < (int)maxb; i++)
      if (K & (1 << i)) mkpath(0, L - spl[i], spn[i]);
  }
  for (int i = 0; i < (int)N; i++) {
  }
  printf("%d\n", N);
  for (int i = 0; i < (int)N; i++) {
    for (int j = 0; j < (int)N; j++) printf("%c", adj[i][j]);
    printf("\n");
  }
}

```