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>
#define MP make_pair
#define PB push_back
#define int long long
#define st first
#define nd second
#define rd third
#define FOR(i, a, b) for(int i =(a); i <=(b); ++i)
#define RE(i, n) FOR(i, 1, n)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define REP(i, n) for(int i = 0;i <(n); ++i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define __builtin_ctz __builtin_ctzll
#define __builtin_clz __builtin_clzll
#define __builtin_popcount __builtin_popcountll
using namespace std;
template<typename TH> void _dbg(const char* sdbg, TH h) { cerr<<sdbg<<"="<<h<<endl; }
template<typename TH, typename... TA> void _dbg(const char* sdbg, TH h, TA... t) {
  while(*sdbg != ',') { cerr<<*sdbg++; } cerr<<"="<<h<<","; _dbg(sdbg+1, t...);
}
#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#define debugv(x) {{cerr <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<endl; }}
#else
#define debug(...) (__VA_ARGS__)
#define debugv(x)
#define cerr if(0)cout
#endif
#define next ____next
#define prev ____prev
#define left ____left
#define hash ____hash
typedef long long ll;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<pair<int, int> > VPII;
typedef vector<pair<ll, ll> > VPLL;

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
template<class T1, class T2>
ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class A, class B, class C> struct Triple { A first; B second; C third;
  bool operator<(const Triple& t) const { if (st != t.st) return st < t.st; if (nd != t.nd) return nd < t.nd; return rd < t.rd; } };
template<class T> void ResizeVec(T&, vector<int>) {}
template<class T> void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]); sz.erase(sz.begin()); if (sz.empty()) { return; }
  for (T& v : vec) { ResizeVec(v, sz); }
}
typedef Triple<int, int, int> TIII;
template<class A, class B, class C>
ostream& operator<< (ostream &out, Triple<A, B, C> t) { return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")"; }
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class T> ostream& operator<<(ostream& out, set<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class L, class R> ostream& operator<<(ostream& out, map<L, R> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }

struct Sol {
  void Test() {
    int n, c, q;
    cin>>n>>c>>q;
    VI p(n + 2);
    RE (i, n) {
      cin>>p[i];
    }
    VVI dp(n + 2, VI(c + 2));
    vector<VVI> jumps(n + 2, VVI(c + 2, VI(20, 2e18)));
    dp[0] = VI(c + 2, 1);
    RE (i, n) {
      REP (j, c + 1) {
        REP (sub_c, j + 1) {
          if (i - 1 - sub_c < 0) { continue; }
          dp[i][j] += dp[i - 1 - sub_c][j - sub_c];
        }
      }
    }
    debug(dp[n]);
    
    RE (i, n) {
      FOR (j, 0, c) {
        int omit = 0;
        RE (steps, j) {
          if (i + steps > n) { continue; }
          if (p[i + steps] < p[i]) {
            omit += dp[n - i - steps][j - steps];
          }
        }
        jumps[i][j][0] = omit;
        //debug(i, j, jumps[i][j][0]);
      }
    }
    cerr<<endl;
    
    RE (layer, 20) {
      RE (i, n) {
        int dist = 1 << (layer - 1);
        if (i + dist > n) { continue; }
        FOR (j, 0, c) {
          jumps[i][j][layer] = jumps[i][j][layer - 1] + jumps[i + dist][j][layer - 1];
          if (jumps[i][j][layer] > 1e18) {
            jumps[i][j][layer] = 1e18;
          }
        }
      }
    }
    
    RE (i, q) {
      int pos, to_omit;
      cin>>pos>>to_omit;
      to_omit--;
      if (to_omit >= dp[n][c]) {
        cout<<"-1\n";
        continue;
      }
      int cur_i = 1;
      int cur_c = c;
      int huh = 0;
      while (1) {
        int layer = 18;
        while (layer >= 0) {
          int nxt = cur_i + (1 << layer);
          if (nxt > n) {
            layer--;
            continue;
          }
          //debug(cur_i, cur_c, layer, jumps[cur_i][cur_c][layer]);
          if (jumps[cur_i][cur_c][layer] <= to_omit && dp[n + 1 - nxt][cur_c] > to_omit - jumps[cur_i][cur_c][layer]) {
            to_omit -= jumps[cur_i][cur_c][layer];
            cur_i = nxt;
            //debug("go", layer);
          }
          layer--;
        }
        if (cur_i > pos) {
          cout<<p[pos]<<"\n";
          goto End;
        }
        // aktualnie mam dp[n + 1 - cur_i][cur_c] mozliwosci
        VPII branches;
        int control = 0;
        FOR (steps, 0, cur_c) {
          if (cur_i + steps > n) { continue; }
          branches.PB({cur_i + steps, dp[n - cur_i - steps][cur_c - steps]});
          control += branches.back().nd;
        }
        //assert(control == dp[n + 1 - cur_i][cur_c]);
        sort(ALL(branches), [&](PII A, PII B) { return p[A.st] < p[B.st]; });
        //debug(to_omit);
        //debug(branches);
        huh++;
        assert(huh <= 6);
        for (auto branch : branches) {
          if (branch.nd <= to_omit) {
            to_omit -= branch.nd;
            continue;
          }
          if (pos <= branch.st) {
            int from_end = pos - cur_i;
            debug(n, from_end, branch.st);
            cout<<p[branch.st - from_end]<<"\n";
            goto End;
          }
          assert(branch.st != cur_i);
          cur_c -= branch.st - cur_i;
          cur_i = branch.st + 1;
          debug("manual", cur_i, cur_c);
          break;
        }
      }
      End: ;
    }
    
    
    
    
    
    
    
  }
};

int32_t main() {

  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  cin.tie(0);
  //double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  
  int t;
  cin>>t;
  RE (i, t) {
    Sol sol;
    sol.Test();
  }
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  return 0;
}