1+ /*
2+
3+ Problem : E - Advertising Agency 1475
4+
5+ Approach : find no of element in which we have choice then apply nCr using MOD
6+
7+ Time Complexity : O(nlogn)
8+ Space Complexity : O(n)
9+
10+ https://codeforces.com/contest/1475/submission/355454137
11+ */
12+ #include < bits/stdc++.h>
13+ #include < bits/stdc++.h>
14+ #include < ext/pb_ds/assoc_container.hpp>
15+ #include < ext/pb_ds/tree_policy.hpp>
16+ using namespace std ;
17+ using namespace __gnu_pbds ;
18+
19+ #define fast_io ios::sync_with_stdio (false ); cin.tie(nullptr );
20+ #define int long long
21+ #define ull unsigned long long
22+ #define ld long double
23+ #define vi vector<int >
24+ #define vb vector<bool >
25+ #define vs vector<string>
26+ #define pii pair<int ,int >
27+ #define pll pair<long long ,long long >
28+ #define vpi vector<pii>
29+ #define vpl vector<pll>
30+ #define vvi vector<vi>
31+ #define mii map<int ,int >
32+ #define si set<int >
33+ #define osi ordered_set<int >
34+ #define loop (i,n ) for (int i = 0 ; i < n; i++)
35+ #define rloop (i,n ) for (int i = n - 1 ; i >= 0 ; i--)
36+ #define loop1 (i,a,b ) for (int i = a; i <= b; i++)
37+ #define rloop1 (i,a,b ) for (int i = a; i >= b; i--)
38+ #define pb push_back
39+ #define ppb pop_back
40+ #define mp make_pair
41+ #define F first
42+ #define S second
43+ #define lb lower_bound
44+ #define ub upper_bound
45+ #define all (x ) (x).begin(), (x).end()
46+ #define rall (x ) (x).rbegin(), (x).rend()
47+ #define sz (x ) ((int )(x).size())
48+ #define findbo (x ) find_by_order(x) // returns iterator to the k-th largest element (0-based)
49+ #define orderbk (x ) order_of_key(x) // returns number of items strictly smaller than x
50+ #define endl ' \n '
51+ #define printy cout<<" YES" <<endl
52+ #define printn cout<<" NO" <<endl
53+ #define vout (a ) for (int i = 0 ; i < a.size(); i++) cout << a[i] << ' ' ; cout << endl;
54+ #define vpout (a ) loop(i,sz(a)) cout <<a[i].F<<' ' << a[i].S<< endl; cout << endl;
55+ #define rt (x ){ cout<<x<<endl; return ; }
56+ template <typename T>
57+ using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; // less_equal<T> for multiset, greater<T> for decreasing order
58+
59+ const int MOD = 1e9 + 7 ;
60+ const int INF = 1e9 ;
61+
62+ #ifndef ONLINE_JUDGE
63+ #define debug (x ) cerr<<#x<<" = " ; _print(x); cerr<<endl;
64+ template <typename T> void _print (const T& x){ cerr<<x; }
65+ template <typename T,typename U> void _print (const pair<T,U>& p){ cerr<<" (" ; _print (p.first ); cerr<<" , " ; _print (p.second ); cerr<<" )" ; }
66+ template <typename T> void _print (const vector<T>& v){ cerr<<" [" ; for (size_t i=0 ;i<v.size ();++i){ _print (v[i]); if (i+1 !=v.size ()) cerr<<" , " ; } cerr<<" ]" ; }
67+ template <typename T> void _print (const set<T>& s){ cerr<<" {" ; for (auto it=s.begin (); it!=s.end (); ++it){ _print (*it); if (next (it)!=s.end ()) cerr<<" , " ; } cerr<<" }" ; }
68+ template <typename T,typename U> void _print (const map<T,U>& m){ cerr<<" {" ; for (auto it=m.begin (); it!=m.end (); ++it){ _print (it->first ); cerr<<" : " ; _print (it->second ); if (next (it)!=m.end ()) cerr<<" , " ; } cerr<<" }" ; }
69+ #else
70+ #define debug (x )
71+ #endif
72+
73+ #define sumV (a ) accumulate(all(a), 0LL )
74+ #define minV (a ) min_element(all(a))
75+ #define maxV (a ) max_element(all(a))
76+
77+ int log2f (long long n) {
78+ if (n==0 ) return -1 ;
79+ int res = 0 ;
80+ while (n > 1 ){
81+ n >>= 1 ;
82+ res++;
83+ }
84+ return res;
85+ }
86+
87+ int binpow (int a, int n) {
88+ int result = 1 ;
89+ a %= MOD;
90+ while (n > 0 ) {
91+ if (n & 1 )
92+ result = (result * a) % MOD;
93+ a = (a * a) % MOD;
94+ n >>= 1 ;
95+ }
96+ return result;
97+ }
98+
99+ int modinv (int a){
100+ return binpow (a,MOD-2 );
101+ }
102+
103+ int nCr (int n,int r) {
104+ if (r<0 || r>n) return 0 ;
105+ if (r>n-r) r = n-r;
106+
107+ int res = 1 ;
108+ for (int i = 1 ; i<=r; i++){
109+ res = (res*(n-r+i))%MOD;
110+ res = (res * modinv (i))%MOD;
111+ }
112+ return res;
113+ }
114+
115+
116+ void solve (){
117+ int n,k;
118+ cin>>n>>k;
119+ vi a (n);
120+ map<int ,int ,greater<int >> mp;
121+ loop (i,n){
122+ cin>>a[i];
123+ mp[a[i]]++;
124+ }
125+
126+ int m;
127+ for (auto it : mp){
128+ if (it.S >k){
129+ m = it.S ;
130+ break ;
131+ }
132+ else k-=it.S ;
133+ }
134+
135+ cout<< nCr (m,k) <<endl;
136+
137+ }
138+
139+ int32_t main (){
140+ fast_io;
141+ int t=1 ;
142+ cin>>t;
143+ while (t--){
144+ solve ();
145+ }
146+ return 0 ;
147+ }
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