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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using intpair = pair<int, int>;
using llpair = pair<ll, ll>;
using vint = vector<int>;
using vll = vector<ll>;
bool ans[1000][1000];
int maxPath;
int lastV = 2;
vint bits;
int gk;
void setEdge(int a, int b) {
  ans[a][b] = true;
  ans[b][a] = true;
}
void setOne() {
  if (maxPath == 1) {
    setEdge(0, 1);
    return;
  }
  setEdge(0, lastV);
  int a = lastV;
  lastV++;
  for (int i = 2; i < maxPath; i++) {
    setEdge(a, lastV);
    a = lastV;
    lastV++;
  }
  setEdge(a, 1);
}
void setTwo() {
  setEdge(0, lastV);
  setEdge(0, lastV + 1);
  if (maxPath == 2) {
    setEdge(1, lastV);
    setEdge(1, lastV + 1);
    lastV += 2;
    return;
  }
  if (bits[0] == 0) {
    setEdge(lastV, 3);
    setEdge(lastV + 1, 3);
    lastV += 2;
    return;
  }
  setEdge(lastV, lastV + 2);
  setEdge(lastV + 1, lastV + 2);
  int a = lastV + 2;
  lastV += 3;
  for (int i = 3; i < maxPath; i++) {
    setEdge(a, lastV);
    a = lastV;
    lastV++;
  }
  setEdge(a, 1);
}
void SetNum(int bit) {
  setEdge(0, lastV);
  setEdge(0, lastV + 1);
  int a = lastV, b = lastV + 1;
  lastV += 2;
  for (int i = 1; i < bit; i++) {
    setEdge(a, lastV);
    setEdge(a, lastV + 1);
    setEdge(b, lastV);
    setEdge(b, lastV + 1);
    a = lastV;
    b = lastV + 1;
    lastV += 2;
  }
  if (maxPath == bit + 1) {
    setEdge(1, a);
    setEdge(1, b);
    return;
  }
  if (bits[0] == bit) {
    setEdge(a, lastV);
    setEdge(b, lastV);
    a = lastV;
    lastV++;
    for (int i = bit + 2; i < maxPath; i++) {
      setEdge(a, lastV);
      a = lastV;
      lastV++;
    }
    setEdge(1, a);
    return;
  }
  if (bits[0] == 0) {
    setEdge(a, bit + 2 * bits[0] + 2);
    setEdge(b, bit + 2 * bits[0] + 2);
  } else {
    setEdge(a, bit - bits[0] + 2 * bits[0] + 2);
    setEdge(b, bit - bits[0] + 2 * bits[0] + 2);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int bit = 0;
  int k;
  cin >> k;
  while (k) {
    if (k % 2 == 1) {
      bits.push_back(bit);
    }
    k >>= 1;
    bit++;
  }
  maxPath = *bits.rbegin() + 1;
  for (auto x : bits) {
    if (x == 0)
      setOne();
    else if (x == 1)
      setTwo();
    else
      SetNum(x);
  }
  cout << lastV << endl;
  for (int i = 0; i < lastV; i++) {
    for (int j = 0; j < lastV; j++) {
      if (ans[i][j])
        cout << "Y";
      else
        cout << "N";
    }
    cout << endl;
  }
}