Problem: Acacius is studying strings theory. Today he came with the following problem.

You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?

Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".

Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 ≤ i ≤ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.

Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.

Input

First line of input contains an integer T (1 ≤ T ≤ 5000), number of test cases. T pairs of lines with test case descriptions follow.

The first line of a test case description contains a single integer n (7 ≤ n ≤ 50), length of a string s.

The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.

Output

For each test case output an answer for it.

In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".

Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.

You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).

Example

Input


6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba


Output


Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No

Note

In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.

In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.

In sixth example there are two occurrences of a string "abacaba" as a substring.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
const int N = 5e5 + 5;
int M = 998244353;
long long fact[N], inv[N];
vector<bool> prime;
vector<int> drow4{1, -1, 0, 0};
vector<int> dcol4{0, 0, -1, 1};
vector<int> drow8{0, 1, 1, 1, 0, -1, -1, -1};
vector<int> dcol8{1, 1, 0, -1, -1, -1, 0, 1};
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << "debug statement : " << *it << " = " << a << endl;
  err(++it, args...);
}
long long powerModM(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % M;
    y = y >> 1;
    x = (x * x) % M;
  }
  return res % M;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
void preCalcFactorial(int n) {
  fact[0] = inv[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % M;
    inv[i] = powerModM(fact[i], M - 2);
  }
}
long long nCr(long long n, long long r) {
  long long ans = ((fact[n] * inv[r]) % M + M) % M;
  ans = ((ans * inv[n - r]) % M + M) % M;
  return ans;
}
void sieve(int n) {
  prime.resize(n + 1, true);
  prime[1] = false;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
class BIT {
 public:
  int size;
  vector<int> v;
  BIT(int second) {
    size = second;
    v.resize(second, 0);
  }
  void update(int i, int delta) {
    while (i < size) {
      v[i] += delta;
      i += i & (-i);
    }
  }
  int sum(int i) {
    int second = 0;
    while (i > 0) {
      second += v[i];
      i -= i & (-i);
    }
    return second;
  }
};
int findIn(string second) {
  string pat = "abacaba";
  if (second.find(pat) == string::npos) return 0;
  int idx = second.find(pat);
  second = second.substr(idx + 1);
  if (second.find(pat) == string::npos) return 1;
  return 2;
}
string fill(string pat, string second, int i) {
  string res = second;
  for (int j = i, k = 0; j < second.length() && k < pat.length(); j++, k++) {
    if (res[j] == '?') {
      res[j] = pat[k];
    }
  }
  return res;
}
void testCase() {
  int n;
  cin >> n;
  string second;
  cin >> second;
  int val = findIn(second);
  if (val == 2) {
    cout << "NO" << endl;
    return;
  }
  if (val == 1) {
    for (auto& i : second) {
      if (i == '?') i = 'z';
    }
    cout << "YES" << endl;
    cout << second << endl;
    return;
  } else {
    string pat = "abacaba";
    for (int i = 0; i < n; i++) {
      if (second[i] == '?' || second[i] == 'a') {
        string temp = fill(pat, second, i);
        int val = findIn(temp);
        if (val == 1) {
          for (auto& i : temp) {
            if (i == '?') i = 'z';
          }
          cout << "YES" << endl;
          cout << temp << endl;
          return;
        }
      }
    }
  }
  cout << "NO" << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    testCase();
  }
}