Problem:
Score : 700 points

Problem Statement

You are given four integers: H, W, h and w (1 ≤ h ≤ H, 1 ≤ w ≤ W).
Determine whether there exists a matrix such that all of the following conditions are held, and construct one such matrix if the answer is positive:

The matrix has H rows and W columns.

Each element of the matrix is an integer between -10^9 and 10^9 (inclusive).

The sum of all the elements of the matrix is positive.

The sum of all the elements within every subrectangle with h rows and w columns in the matrix is negative.

Constraints

1 ≤ h ≤ H ≤ 500

1 ≤ w ≤ W ≤ 500

Input

Input is given from Standard Input in the following format:

H W h w

Output

If there does not exist a matrix that satisfies all of the conditions, print No.

Otherwise, print Yes in the first line, and print a matrix in the subsequent lines in the following format:

a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}

Here, a_{ij} represents the (i,\ j) element of the matrix.

Sample Input 1

3 3 2 2

Sample Output 1

Yes
1 1 1
1 -4 1
1 1 1

The sum of all the elements of this matrix is 4, which is positive.
Also, in this matrix, there are four subrectangles with 2 rows and 2 columns as shown below. For each of them, the sum of all the elements inside is -1, which is negative.

Sample Input 2

2 4 1 2

Sample Output 2

No

Sample Input 3

3 4 2 3

Sample Output 3

Yes
2 -5 8 7
3 -5 -4 -5
2 1 -1 7

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <vector>

const int MOD = 1e9 + 7;
const int iINF = 2147483647 / 2;
const long long int llINF = 9223372036854775807 / 2;

using namespace std;
using ll = long long int;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;

typedef pair<ll, ll> pll;
bool paircomp(const pll &a, const pll &b) {
  if (a.first == b.first)
    return a.second < b.second;
  return a.first < b.first;
}

#define REP(i, n) for (ll i = 0; i < (n); i++)
#define RREP(i, n) for (ll i = (n)-1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define AUTO(i, m) for (auto &i : m)
#define ALL(a) (a).begin(), (a).end()
#define MAX(vec) *std::max_element(vec.begin(), vec.end())
#define MIN(vec) *std::min_element(vec.begin(), vec.end())
#define ARGMAX(vec)                                                            \
  std::distance(vec.begin(), std::max_element(vec.begin(), vec.end()))
#define ARGMIN(vec)                                                            \
  std::distance(vec.begin(), std::min_element(vec.begin(), vec.end()))
#define REV(T) greater<T>()
#define PQ(T) priority_queue<T, vector<T>, greater<T>>
#define VVL(a, b, c) vector<vector<ll>>(a, vector<ll>(b, c))
#define VV(T, a, b, c) vector<vector<T>>(a, vector<T>(b, c))
#define VVVL(a, b, c, d)                                                       \
  vector<vector<vector<ll>>>(a, vector<vector<ll>>(b, vector<ll>(c, d)))
#define VVV(T, a, b, c, d)                                                     \
  vector<vector<vector<T>>>(a, vector<vector<T>>(b, vector<T>(c, d)))
#define SP(a) setprecision(a)
#define SQRT(a) sqrt((long double)(a))
#define DPOW(a, b) pow((long double)(a), (long double)(b))
#define UNIQUE(vec)                                                            \
  do {                                                                         \
    sort(ALL((vec)));                                                          \
    (vec).erase(std::unique(ALL((vec))), (vec).end());                         \
  } while (0)

ll POW(ll n, ll m) {
  if (m == 0) {
    return 1;
  } else if (m % 2 == 0) {
    ll tmp = POW(n, m / 2);
    return (tmp * tmp);
  } else {
    return (n * POW(n, m - 1));
  }
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll H, W, h, w;
  cin >> H >> W >> h >> w;
  ll poyo = (H / h) * (W / w);
  if ((H * W - poyo) * 100 - ((poyo * (h * w - 1) * 100) + 1) > 0) {
    cout << "Yes" << endl;
    FOR(i, 1, H + 1) {
      FOR(j, 1, W + 1) {
        if (i % h == 0 && j % w == 0)
          cout << -(h * w - 1) * 100 - 1 << " ";
        else
          cout << 100 << " ";
      }
      cout << endl;
    }
  } else
    cout << "No" << endl;

  return 0;
}