C. Perfect Keyboard

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

Polycarp uses the same password

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

Input

The first line contains one integer

Then

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

Example

input

Copy

5
ababa
codedoca
abcda
zxzytz

abcdefghijklmnopqrstuvwxyza

output

Copy

YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO


有两种情况不可行：一个字母相邻字符数大于2的不可以；构成环的不可以
符合的就是一条链
从链的端点开始输出即可。

#include <bits/stdc++.h>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const int mod = 998244353;
const int N = 1e5+5;
inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m%n);
}
ll lcm(ll m, ll n)
{
    return m*n / gcd(m, n);
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        string s;
        cin>>s;
        map<char,bool> vis;
        for(int i=0;i<26;i++)
        {
            vis[i+'a']=0;
        }
        map<char,set<char>> mp;
        int fg=0;
        for(int i=0;i<s.size();i++)
        {
            if(i-1>=0)
                mp[s[i]].insert(s[i-1]);
            if(i+1<s.size())
                mp[s[i]].insert(s[i+1]);
            if(mp[s[i]].size()>2)
            {
                fg=1;
                break;
            }
        }
        if(fg)
        {
            cout<<"NO"<<endl;
            continue;
        } else{
            string res;

            char st='A';
            for(auto i:mp)
            {
                if(i.second.size()<=1)
                {
                    st=i.first;
                    break;
                }
            }
            if(st=='A')
            {
                cout<<"NO"<<endl;
                continue;
            }
            cout<<"YES"<<endl;
            vis[st]=1;
            for(int i=0;i<mp.size();i++)
            {
                cout<<st;
                for(auto j:mp[st])
                {
                    if(!vis[j])
                    {
                        st=j;
                        vis[st]=1;
                    }
                }
            }
            for(auto i:vis)
            {
                if(i.second==0)
                    cout<<i.first;
            }
            cout<<endl;
        }
    }
    return 0;
}