Problem: Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all 26 lowercase Latin letters will be arranged in some order.

Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in s, so, for example, the password cannot be password (two characters s are adjacent).

Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?

Input

The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases.

Then T lines follow, each containing one string s (1 ≤ |s| ≤ 200) representing the test case. s consists of lowercase Latin letters only. There are no two adjacent equal characters in s.

Output

For each test case, do the following:

  * if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem); 
  * otherwise, print YES (in upper case), and then a string consisting of 26 lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them. 

Example

Input


5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza


Output


YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) { return n | (1LL << i); }
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx4[] = {0, 0, +1, -1};
int dy4[] = {+1, -1, 0, 0};
int dx8[] = {+1, 0, -1, 0, +1, +1, -1, -1};
int dy8[] = {0, +1, 0, -1, +1, -1, +1, -1};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline bool isIntege(double num) { return (num == (int)num); }
inline bool collinear(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2);
}
inline double coDist(double x1, double y1, double x2, double y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
inline double TriangleAreaWithSide(double a, double b, double c) {
  double s = (a + b + c) / 2;
  double area = sqrt(s * (s - a) * (s - b) * (s - c));
  return area;
}
inline double area3(double x1, double y1, double x2, double y2, double x3,
                    double y3) {
  double A = abs((x1 * y2 + x2 * y3 + x3 * y1) - (y1 * x2 + y2 * x3 + y3 * x1));
  A /= 2.0;
  return A;
}
inline long double degreetoradian(long double x) {
  long double val = PI * x;
  val /= (180.0);
  return val;
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
template <typename T>
inline T gcd(T a, T b) {
  T c;
  while (b) {
    c = b;
    b = a % b;
    a = c;
  }
  return a;
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
inline bool equalTo(double a, double b) {
  if (fabs(a - b) <= eps)
    return true;
  else
    return false;
}
inline bool notEqual(double a, double b) {
  if (fabs(a - b) > eps)
    return true;
  else
    return false;
}
inline bool lessThan(double a, double b) {
  if (a + eps < b)
    return true;
  else
    return false;
}
inline bool lessThanEqual(double a, double b) {
  if (a < b + eps)
    return true;
  else
    return false;
}
inline bool greaterThan(double a, double b) {
  if (a > b + eps)
    return true;
  else
    return false;
}
inline bool greaterThanEqual(double a, double b) {
  if (a + eps > b)
    return true;
  else
    return false;
}
inline string to_s(int t) {
  stringstream ss;
  ss << t;
  return ss.str();
}
struct edge {
  int p, q, w;
};
bool cmp(edge &a, edge &b) { return a.w < b.w; }
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int vis[30][30], vis1[30];
vector<int> v[30];
vector<int> ans;
void dfs(int u) {
  vis1[u] = 1;
  ans.emplace_back(u);
  for (int i = 0; i < v[u].size(); i++) {
    if (vis1[v[u][i]] == 0) {
      dfs(v[u][i]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    if (s.size() == 1) {
      cout << "YES" << '\n';
      for (char c = 'a'; c <= 'z'; c++) cout << c;
      cout << '\n';
      continue;
    }
    for (int i = 0; i < s.length() - 1; i++) {
      if (vis[s[i] - 'a'][s[i + 1] - 'a'] == 0) {
        v[s[i] - 'a'].emplace_back(s[i + 1] - 'a');
        v[s[i + 1] - 'a'].emplace_back(s[i] - 'a');
        vis[s[i] - 'a'][s[i + 1] - 'a'] = 1;
        vis[s[i + 1] - 'a'][s[i] - 'a'] = 1;
      }
    }
    int id = -1, ok = 1;
    for (int i = 0; i < 26; i++) {
      if (v[i].size() == 1) id = i;
      if (v[i].size() > 2) {
        ok = 0;
        break;
      }
    }
    if (id == -1 || !ok) {
      cout << "NO" << '\n';
      memset(vis, 0, sizeof(vis));
      memset(vis1, 0, sizeof(vis1));
      for (int i = 0; i < 26; i++) {
        v[i].clear();
      }
      continue;
    }
    dfs(id);
    for (int i = 0; i < 26; i++) {
      if (vis1[i] == 0) ans.emplace_back(i);
      v[i].clear();
    }
    cout << "YES" << '\n';
    for (int i = 0; i < ans.size(); i++) cout << (char)(ans[i] + 'a');
    cout << '\n';
    memset(vis, 0, sizeof(vis));
    memset(vis1, 0, sizeof(vis1));
    ans.clear();
  }
}