#include <cstdio>
#include <vector>
#include <algorithm> 
using namespace std;
const int B = 1000;	//桶的大小
const int MAX_N = 100000;
const int MAX_M = 5000;
//输入 
int N, M;
int A[MAX_N];
int I[MAX_M], J[MAX_M], K[MAX_M];

int nums[MAX_N]; //对A排序之后的结果 
vector<int > bucket[MAX_N / B];	//每个桶排序之后的结果 

void solve(){
	for(int i = 0;i < N;i ++){
		bucket[i / B].push_back(A[i]);
		nums[i] = A[i];
	}
	sort(nums, nums + N);
	//虽然每B个一组剩下的部分所在的桶没有排序，但是不会产生问题
	for(int i = 0;i < N / B;i ++){
		sort(bucket[i].begin(), bucket[i].end());
	}
	for(int i = 0;i < M;i ++){	//求[l, r)区间中第k个数 
		int l = I[i] - 1, r = J[i], k = K[i];
		int lb = -1, ub = N - 1;
		while(ub - lb > 1){	//二分查找 
			int md = (lb + ub) / 2;
			int x = nums[md];
			int tl = l, tr = r, c = 0;
			//区间两端多出的部分
			while(tl < tr && tl % B != 0){
				if(A[tl ++] <= x){
					c ++;
				}
			} 
			while (tl < tr && tr % B != 0){
				if(A[-- tr] <= x){
					c ++;
				}
			} 
			//对每一个桶进行计算
			while(tl < tr){
				int b = tl / B;
				c += upper_bound(bucket[b].begin(), bucket[b].end(), x)
					 - bucket[b].begin();
				tl += B;
			} 
			if(c >= k){
				ub = md;
			}else{
				lb = md;
			}
		}
		printf("%d\n", nums[ub]);
	}
}

int main(){
	int T, x;
	scanf("%d%d", &N, &M);
	for(int i = 0;i < N;i ++){
		scanf("%d", &A[i]);
	}
	for(int i = 0;i < M;i ++){
		scanf("%d%d%d", &I[i], &J[i], &K[i]);
	}
	solve();
	return 0;
}