You are given four positive integers n, m, a, b (1 ≤ b ≤ n ≤ 50; 1 ≤ a ≤ m ≤ 50). Find any such rectangular matrix of size n × m that satisfies all of the following conditions:

  * each row of the matrix contains exactly a ones; 
  * each column of the matrix contains exactly b ones; 
  * all other elements are zeros. 



If the desired matrix does not exist, indicate this.

For example, for n=3, m=6, a=2, b=1, there exists a matrix satisfying the conditions above:

$$$ \begin{vmatrix} 0 & 1 & 0 & 0 & 0 & 1 \\\ 1 & 0 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 0 \end{vmatrix} $$$

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

Each test case is described by four positive integers n, m, a, b (1 ≤ b ≤ n ≤ 50; 1 ≤ a ≤ m ≤ 50), where n and m are the sizes of the matrix, and a and b are the number of ones for rows and columns, respectively.

Output

For each test case print:

  * "YES" (without quotes) and the required matrix (if there are several answers, print any) if it exists, or 
  * "NO" (without quotes) if it does not exist. 



To print the matrix n × m, print n rows, each of which consists of m numbers 0 or 1 describing a row of the matrix. Numbers must be printed without spaces.

Example

Input


5
3 6 2 1
2 2 2 1
2 2 2 2
4 4 2 2
2 1 1 2


Output


YES
010001
100100
001010
NO
YES
11
11
YES
1100
1100
0011
0011
YES
1
1
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("-ffloat-store")
const long long INF = 9e18;
const long long N = 1e9 + 7;
const double eps = 1e-9;
const auto start_time = std::chrono::high_resolution_clock::now();
void zark() {}
void solve() {
  long long n, m, a, b;
  cin >> n >> m >> a >> b;
  vector<vector<long long>> ans(n, vector<long long>(m));
  vector<long long> r(n);
  for (long long j = (0); j < (m); j++) {
    vector<pair<long long, long long>> rr(n);
    for (long long i = (0); i < (n); i++) rr[i].first = r[i], rr[i].second = i;
    sort((rr).begin(), (rr).end());
    for (long long i = (0); i < (b); i++)
      ans[rr[i].second][j] = 1, r[rr[i].second]++;
  }
  bool ok = true;
  for (long long i = (0); i < (n); i++) {
    long long s = 0;
    for (long long j = (0); j < (m); j++) s += ans[i][j];
    ok &= s == a;
  }
  for (long long j = (0); j < (m); j++) {
    long long s = 0;
    for (long long i = (0); i < (n); i++) s += ans[i][j];
    ok &= s == b;
  }
  if (ok) {
    cout << "YES\n";
    for (long long i = (0); i < (n); i++) {
      for (long long j = (0); j < (m); j++) cout << ans[i][j];
      cout << "\n";
    }
  } else
    cout << "NO\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
  zark();
  return 0;
}
