Problem: Alice had a permutation p_1, p_2, …, p_n. Unfortunately, the permutation looked very boring, so she decided to change it and choose some non-overlapping subranges of this permutation and reverse them. The cost of reversing a single subrange [l, r] (elements from position l to position r, inclusive) is equal to r - l, and the cost of the operation is the sum of costs of reversing individual subranges. Alice had an integer c in mind, so she only considered operations that cost no more than c.

Then she got really bored, and decided to write down all the permutations that she could possibly obtain by performing exactly one operation on the initial permutation. Of course, Alice is very smart, so she wrote down each obtainable permutation exactly once (no matter in how many ways it can be obtained), and of course the list was sorted lexicographically.

Now Bob would like to ask Alice some questions about her list. Each question is in the following form: what is the i-th number in the j-th permutation that Alice wrote down? Since Alice is too bored to answer these questions, she asked you to help her out.

Input

The first line contains a single integer t (1 ≤ t ≤ 30) — the number of test cases.

The first line of each test case contains three integers n, c, q (1 ≤ n ≤ 3 ⋅ 10^4, 1 ≤ c ≤ 4, 1 ≤ q ≤ 3 ⋅ 10^5) — the length of the permutation, the maximum cost of the operation, and the number of queries.

The next line of each test case contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j if i ≠ j), describing the initial permutation.

The following q lines describe the queries. Each of them contains two integers i and j (1 ≤ i ≤ n, 1 ≤ j ≤ 10^{18}), denoting parameters of this query.

It is guaranteed that the sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values q over all test cases does not exceed 3 ⋅ 10^5.

Output

For each query output the answer for this query, or -1 if j-th permutation does not exist in her list.

Examples

Input


2
3 1 9
1 2 3
1 1
2 1
3 1
1 2
2 2
3 2
1 3
2 3
3 3
6 4 4
6 5 4 3 1 2
1 1
3 14
1 59
2 6


Output


1
2
3
1
3
2
2
1
3
1
4
-1
5


Input


1
12 4 2
1 2 3 4 5 6 7 8 9 10 11 12
2 20
2 21


Output


2
2

Note

In the first test case, Alice wrote down the following permutations: [1, 2, 3], [1, 3, 2], [2, 1, 3].

Note that, for a permutation [3, 2, 1] Alice would have to reverse the whole array, and it would cost her 2, which is greater than the specified value c=1. The other two permutations can not be obtained by performing exactly one operation described in the problem statement.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#define si(x) int(x.size())
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> bool chmax(t&a,u b){if(a<b){a=b;return true;}else return false;}
template<class t,class u> bool chmin(t&a,u b){if(b<a){a=b;return true;}else return false;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

#define mp make_pair
#define mt make_tuple
#define one(x) memset(x,-1,sizeof(x))
#define zero(x) memset(x,0,sizeof(x))
#ifdef LOCAL
void dmpr(ostream&os){os<<endl;}
template<class T,class... Args>
void dmpr(ostream&os,const T&t,const Args&... args){
	os<<t<<" ";
	dmpr(os,args...);
}
#define dmp2(...) dmpr(cerr,__LINE__,##__VA_ARGS__)
#else
#define dmp2(...) void(0)
#endif

using uint=unsigned;
using ull=unsigned long long;

template<class t,size_t n>
ostream& operator<<(ostream&os,const array<t,n>&a){
	return os<<vc<t>(all(a));
}

template<int i,class T>
void print_tuple(ostream&,const T&){
}

template<int i,class T,class H,class ...Args>
void print_tuple(ostream&os,const T&t){
	if(i)os<<",";
	os<<get<i>(t);
	print_tuple<i+1,T,Args...>(os,t);
}

template<class ...Args>
ostream& operator<<(ostream&os,const tuple<Args...>&t){
	os<<"{";
	print_tuple<0,tuple<Args...>,Args...>(os,t);
	return os<<"}";
}

template<class t>
void print(t x,int suc=1){
	cout<<x;
	if(suc==1)
		cout<<"\n";
	if(suc==2)
		cout<<" ";
}

ll read(){
	ll i;
	cin>>i;
	return i;
}

vi readvi(int n,int off=0){
	vi v(n);
	rep(i,n)v[i]=read()+off;
	return v;
}

pi readpi(int off=0){
	int a,b;cin>>a>>b;
	return pi(a+off,b+off);
}

template<class t,class u>
void print(const pair<t,u>&p,int suc=1){
	print(p.a,2);
	print(p.b,suc);
}

template<class T>
void print(const vector<T>&v,int suc=1){
	rep(i,v.size())
		print(v[i],i==int(v.size())-1?suc:2);
}

string readString(){
	string s;
	cin>>s;
	return s;
}

template<class T>
T sq(const T& t){
	return t*t;
}

//#define CAPITAL
void yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<"\n";
	#else
	cout<<"Yes"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}
void no(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<"\n";
	#else
	cout<<"No"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}
void possible(bool ex=true){
	#ifdef CAPITAL
	cout<<"POSSIBLE"<<"\n";
	#else
	cout<<"Possible"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}
void impossible(bool ex=true){
	#ifdef CAPITAL
	cout<<"IMPOSSIBLE"<<"\n";
	#else
	cout<<"Impossible"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}

constexpr ll ten(int n){
	return n==0?1:ten(n-1)*10;
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

int topbit(signed t){
	return t==0?-1:31-__builtin_clz(t);
}
int topbit(ll t){
	return t==0?-1:63-__builtin_clzll(t);
}
int botbit(signed a){
	return a==0?32:__builtin_ctz(a);
}
int botbit(ll a){
	return a==0?64:__builtin_ctzll(a);
}
int popcount(signed t){
	return __builtin_popcount(t);
}
int popcount(ll t){
	return __builtin_popcountll(t);
}
bool ispow2(int i){
	return i&&(i&-i)==i;
}
ll mask(int i){
	return (ll(1)<<i)-1;
}

bool inc(int a,int b,int c){
	return a<=b&&b<=c;
}

template<class t> void mkuni(vc<t>&v){
	sort(all(v));
	v.erase(unique(all(v)),v.ed);
}

ll rand_int(ll l, ll r) { //[l, r]
	#ifdef LOCAL
	static mt19937_64 gen;
	#else
	static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
	#endif
	return uniform_int_distribution<ll>(l, r)(gen);
}

template<class t>
void myshuffle(vc<t>&a){
	rep(i,si(a))swap(a[i],a[rand_int(0,i)]);
}

template<class t>
int lwb(const vc<t>&v,const t&a){
	return lower_bound(all(v),a)-v.bg;
}

vvc<int> readGraph(int n,int m){
	vvc<int> g(n);
	rep(i,m){
		int a,b;
		cin>>a>>b;
		//sc.read(a,b);
		a--;b--;
		g[a].pb(b);
		g[b].pb(a);
	}
	return g;
}

vvc<int> readTree(int n){
	return readGraph(n,n-1);
}

const int nmax=30010;
const int cmax=5;

int dp[cmax][nmax],sum[cmax][nmax];
vi to[nmax];

void slv(){
	int n,c,q;cin>>n>>c>>q;
	c++;
	vi p=readvi(n);
	rep(i,n){
		to[i].resize(min(c,n-i));
		iota(all(to[i]),0);
		sort(all(to[i]),[&](int a,int b){
			return p[i+a]<p[i+b];
		});
	}
	per(k,c){
		dp[k][n]=1;
		per(i,n){
			for(auto j:to[i])if(k+j<c){
				if(j==0)sum[k][i+1]=dp[k][i];
				dp[k][i]+=dp[k+j][i+j+1];
			}
		}
		rep(i,n)sum[k][i+1]+=sum[k][i];
	}
	rep(_,q){
		int pos;cin>>pos;pos--;
		int idx;cin>>idx;idx--;
		if(idx>=dp[0][0]){
			print(-1);
			continue;
		}
		int cur=0,used=0;
		while(1){
			int lw=cur,up=n+1;
			auto check=[&](int i){
				int val=sum[used][i]-sum[used][cur];
				return val<=idx&&idx<val+dp[used][i];
			};
			assert(check(lw));
			while(up-lw>1){
				const int mid=(lw+up)/2;
				if(check(mid))lw=mid;
				else up=mid;
			}
			idx-=sum[used][lw]-sum[used][cur];
			cur=lw;
			if(pos<cur){
				print(p[pos]);
				break;
			}
			int done=0;
			for(auto j:to[cur])if(used+j<c){
				if(idx<dp[used+j][cur+j+1]){
					if(pos<=cur+j){
						print(p[cur+(cur+j-pos)]);
						done=1;
						break;
					}else{
						cur+=j+1;
						used+=j;
						done=2;
						break;
					}
				}else{
					idx-=dp[used+j][cur+j+1];
				}
			}
			assert(done);
			if(done==1)break;
		}
	}
	zero(dp);
	zero(sum);
	rep(i,n)to[i].clear();
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
	
	int t;cin>>t;rep(_,t)
	slv();
}