Problem: You are given three strings s, t and p consisting of lowercase Latin letters. You may perform any number (possibly, zero) operations on these strings.

During each operation you choose any character from p, erase it from p and insert it into string s (you may insert this character anywhere you want: in the beginning of s, in the end or between any two consecutive characters). 

For example, if p is aba, and s is de, then the following outcomes are possible (the character we erase from p and insert into s is highlighted):

  * aba → ba, de → ade; 
  * aba → ba, de → dae; 
  * aba → ba, de → dea; 
  * aba → aa, de → bde; 
  * aba → aa, de → dbe; 
  * aba → aa, de → deb; 
  * aba → ab, de → ade; 
  * aba → ab, de → dae; 
  * aba → ab, de → dea; 



Your goal is to perform several (maybe zero) operations so that s becomes equal to t. Please determine whether it is possible.

Note that you have to answer q independent queries.

Input

The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query is represented by three consecutive lines.

The first line of each query contains the string s (1 ≤ |s| ≤ 100) consisting of lowercase Latin letters.

The second line of each query contains the string t (1 ≤ |t| ≤ 100) consisting of lowercase Latin letters.

The third line of each query contains the string p (1 ≤ |p| ≤ 100) consisting of lowercase Latin letters.

Output

For each query print YES if it is possible to make s equal to t, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


4
ab
acxb
cax
a
aaaa
aaabbcc
a
aaaa
aabbcc
ab
baaa
aaaaa


Output


YES
YES
NO
NO

Note

In the first test case there is the following sequence of operation: 

  1. s =  ab, t =  acxb, p =  cax; 
  2. s =  acb, t =  acxb, p =  ax; 
  3. s =  acxb, t =  acxb, p =  a. 



In the second test case there is the following sequence of operation: 

  1. s =  a, t =  aaaa, p =  aaabbcc; 
  2. s =  aa, t =  aaaa, p =  aabbcc; 
  3. s =  aaa, t =  aaaa, p =  abbcc; 
  4. s =  aaaa, t =  aaaa, p =  bbcc. 

Solution (in CPP):

Solution:
#include<bits/stdc++.h> /*
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including */
using namespace std;
//using namespace __gnu_pbds;

//typedefs

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<int>::iterator vit;
typedef set<int>::iterator sit;

/*
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename F, typename S>
using ordered_map = tree<F, S, less<F>, rb_tree_tag, tree_order_statistics_node_update>;
*/

//#Defines
#define rep(i,a,b)	for(i=a;i<=b;i++)
#define repR(i,a,b)	for(i=a;i>=b;i--)
//#define pb push_back
#define pb emplace_back
#define F first
#define S second
#define mp make_pair
#define all(c)		c.begin(),c.end()
#define endl '\n'
#define pf printf
#define sf scanf
//#define left __left
//#define right __right
//#define tree __tree
#define MOD 1000000007
//#define harmonic(n) 0.57721566490153286l+log(n)

#define RESET(a,b)	memset(a,b,sizeof(a))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const ll infLL = 9000000000000000000;


//Bit Operations
inline bool checkBit(ll n, int i) { return n&(1LL<<i); }
inline ll setBit(ll n, int i) { return n|(1LL<<i);; }
inline ll resetBit(ll n, int i) { return n&(~(1LL<<i)); }

int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};


//Inline functions

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }
inline bool isLeapYear(ll year) { return (year%400==0) || (year%4==0 && year%100!=0); }
inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }
inline bool isInside(pii p,ll n,ll m)	{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m); }
inline bool isInside(pii p,ll n)		{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n); }
inline bool isSquare(ll x) 				{ ll s = sqrt(x);	return (s*s==x); }
inline bool isFib(ll x) 				{ return isSquare(5*x*x+4)|| isSquare(5*x*x-4); }
inline bool isPowerOfTwo(ll x)			{ return ((1LL<<(ll)log2(x))==x); }


struct func
{
	//this is a sample overloading function for sorting stl
	bool operator()(pii const &a, pii const &b)	
	{
		if(a.F==b.F)
			return (a.S<b.S);
		return (a.F<b.F);
	}
};

//Prime Number Generator

/*
#define M 100000000
int marked[M/64 + 2];

#define on(x) (marked[x/64] & (1<<((x%64)/2)))
#define mark(x)  marked[x/64] |= (1<<((x%64)/2))

vl prime;

bool isPrime(int num) {
  return num > 1 && (num == 2 || ((num & 1) && !on(num))); 
}

void sieve(ll n) {
	for (ll i = 3; i * i < n; i += 2) {
		if (!on(i)) {
			for (ll j = i * i; j <= n; j += i + i) {
				mark(j);
			}
		}
	}
	prime.pb(2);
	for(ll i = 3; i <= n; i += 2) {
		if(!on(i))	prime.pb(i);	
	}
}

*/

//
//debug
#ifdef tajir
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & 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 F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << it -> first << " = " << it -> second ;
		}
	return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () {
	cerr << endl;
}

template <typename T>
void faltu( T a[], int n ) {
	for(int i = 0; i < n; ++i) cerr << a[i] << ' ';
	cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) {
	cerr << arg << ' ';
	faltu(rest...);
}
#else
#define dbg(args...)
#endif // tajir

bool isOk(string s, string t) {
	int j = 0;
	for(int i = 0; i < s.size(); ++i) {
		char c = s[i];
		for( ; j < t.size() && c != t[j]; ++j);

		if(j == t.size())
			return false;
		if(c != t[j])
			return false;
		else
			++j;
		dbg(c,j);
	}
	return true;
}

int main()
{
	#ifdef tajir
		freopen("input.txt", "r", stdin);
	#else
		// online submission
	#endif
	optimize();
	
	int T;

	cin >> T;

	while(T--) {
		string s,t,p;

		cin >> s >> t >> p;

		vector < int > cnts(26,0) , cntt(26,0) , cntp(26,0);

		for(int i = 0;i < s.size(); ++i)
			cnts[s[i]-'a']++;
		
		for(int i = 0;i < t.size(); ++i)
			cntt[t[i]-'a']++;
		
		for(int i = 0;i < p.size(); ++i)
			cntp[p[i]-'a']++;
		
		bool flag = true;

		for(int i = 0;i < 26; ++i) {
			if(cnts[i] > cntt[i])
				flag = false;
		}

		for(int i = 0;i < 26; ++i) {
			int x = cnts[i] + cntp[i];

			int y = cntt[i];

			if(x < y)
				flag = false;
		}

		if(!isOk(s,t))
			flag = false;

		if(flag)
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}
	
	return 0;
}






//?