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;

#define mk make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

template<typename T> inline void ckmax(T& x, T y) { x = (y > x ? y : x); }
template<typename T> inline void ckmin(T& x, T y) { x = (y < x ? y : x); }

const int MAXC = 4;
const int MAXN = 3e4;

int n, c, q, a[MAXN + 5];

ll ways_eq(int len, int c) {
	// 长度为 len, 操作代价和恰好为 c
	// comb(len - 1, c)
	if (len <= 1) {
		return c == 0;
	}
	if (len - 1 < c) {
		return 0;
	}
	
	ll res = 1;
	for (int i = len - 1; i >= len - c; --i) {
		res *= i;
	}
	for (int i = c; i > 1; --i) {
		res /= i;
	}
	return res;
}
ll ways_leq(int len, int c) {
	// 长度为 len, 操作代价和小于或等于 c
	ll res = 0;
	for (int i = 0; i <= c; ++i) {
		res += ways_eq(len, i);
	}
	return res;
}

struct Node {
	int l, r;
	ll w; // 后面的操作方案数
	
	Node() {}
	Node(int _l, int _r, ll _w) {
		l = _l;
		r = _r;
		w = _w;
	}
};

Node dq[MAXC + 1][MAXN * MAXC * 2 + 10];
int ql[MAXC + 1], qr[MAXC + 1];
int sum[MAXC + 1][MAXN + 5];
ll sumw[MAXC + 1][MAXN * MAXC + 5];

bool cmp(Node lhs, Node rhs) {
	return a[lhs.r] < a[rhs.r];
}

Node F(int st, int s, ll rank) {
	int i = sum[s][st]; // 在 st 之前的, 一共有这么多区间
	/*
	// 暴力查找
	ll cur = 0;
	for (i = ql[s] + i; i <= qr[s]; ++i) {
		cur += dq[s][i].w;
		assert(cur == sumw[s][i - ql[s] + 1] - sumw[s][sum[s][st]]);
		if (cur >= rank) {
			return Node(dq[s][i].l, dq[s][i].r, cur - dq[s][i].w);
		}
	}
	*/
	
	int l = i + 1, r = qr[s] - ql[s] + 2;
	while (l < r) {
		int mid = (l + r) >> 1;
		if (sumw[s][mid] - sumw[s][i] >= rank) {
			r = mid;
		} else {
			l = mid + 1;
		}
	}
	assert(l <= qr[s] - ql[s] + 1);
	return Node(dq[s][ql[s] + l - 1].l, dq[s][ql[s] + l - 1].r, sumw[s][l - 1] - sumw[s][i]);
}

void solve_case() {
	cin >> n >> c >> q;
	for (int i = 1; i <= n; ++i) {
		cin >> a[i];
	}
	for (int s = 1; s <= c; ++s) { // 总代价小于或等于 s
		
		
//		cerr << "-------- maxcost " << s << " --------" << endl;
		ql[s] = MAXN * MAXC + 5, qr[s] = MAXN * MAXC + 4; // 队列清空
		for (int i = 1; i <= n; ++i)
			sum[s][i] = 0;
		
		dq[s][++qr[s]] = Node(n, n, 1); // 什么都不翻转
		for (int i = n - 1; i >= 1; --i) {
			int dl = 0, dr = 0;
			for (int j = 1; j <= min(s, n - i); ++j) {
				// 翻转区间 [i, i + j]
				ll w = ways_leq(n - (i + j), s - j);
				
				if (a[i + j] < a[i]) {
					// 翻转后是 a[i + j], 不翻转是 a[i], 两者比一比
					// 翻转后更小, push_front
					dq[s][ql[s] - (++dl)] = Node(i, i + j, w);
					sum[s][i + 1]++;
				} else {
					dq[s][qr[s] + (++dr)] = Node(i, i + j, w);
				}
			}
			
			if (dl) {
				sort(dq[s] + ql[s] - dl, dq[s] + ql[s], cmp);
				ql[s] -= dl;
			}
			if (dr) {
				sort(dq[s] + qr[s] + 1, dq[s] + qr[s] + dr + 1, cmp);
				qr[s] += dr;
			}
		}
//		cerr << "print queue: " << endl;
//		for (int i = ql[s]; i <= qr[s]; ++i) {
//			cerr << dq[s][i].l << " " << dq[s][i].r << " " << dq[s][i].w << endl;
//		}
//		cerr << "queue end" << endl;
		
		for (int i = 1; i <= n; ++i) {
			sum[s][i] += sum[s][i - 1];
		}
		for (int i = ql[s]; i <= qr[s]; ++i) {
			sumw[s][i - ql[s] + 1] = sumw[s][i - ql[s]] + dq[s][i].w;
		}
	}
	
	ll lim = ways_leq(n, c);
	for (int tq = 1; tq <= q; ++tq) {
		int pos;
		ll rank;
//		cerr << "-------- query: " << endl;
		cin >> pos >> rank;
		
		if (rank > lim) {
			cout << -1 << endl;
			continue;
		}
		
		vector<pii> revs;
		int p = 1;
		int s = c;
		while (1) {
			Node t = F(p, s, rank);
			
//			cerr << "** " << t.l << " " << t.r << " " << t.w << endl;
			
			revs.push_back(make_pair(t.l, t.r));
			
			assert(t.w < rank);
			rank -= t.w;
			s -= (t.r - t.l);
			p = t.r + 1;
			
			if (!s)
				break;
			if (p > n)
				break;
		}
		
		/*
		// 暴力翻转
		static int aa[MAXN + 5];
		for (int i = 1; i <= n; ++i) {
			aa[i] = a[i];
		}
		for (int i = 0; i < SZ(revs); ++i) {
			reverse(aa + revs[i].fi, aa + revs[i].se + 1);
		}
		cout << aa[pos] << endl;
		*/
		
		bool flag = 0;
		for (int i = 0; i < SZ(revs); ++i) {
			if (revs[i].fi <= pos && revs[i].se >= pos) {
				cout << a[revs[i].se - (pos - revs[i].fi)] << endl;
				flag = 1;
				break;
			}
		}
		if (!flag) {
			cout << a[pos] << endl;
		}
	}
}
int main() {
	int T; cin >> T; while (T--) {
		solve_case();
	}
	return 0;
}