#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

/*==========================================================*\
| 划分树
\*==========================================================*/
#define MAX_SIZE 100005
int sorted[MAX_SIZE];//已经排好序的数据
int toleft[25][MAX_SIZE];
int tree[25][MAX_SIZE];
void build_tree(int left, int right, int deep) 
{
	int i;
	if (left == right) return ;
	int mid = (left + right) >> 1;
	int same = mid - left + 1; //位于左子树的数据
	for (i = left; i <= right; ++i) {//计算放于左子树中与中位数相等的数字个数
		if (tree[deep][i] < sorted[mid]) {
			--same;
		}
	}
	int ls = left;
	int rs = mid + 1;
	for (i = left; i <= right; ++i) {
		int flag = 0;
		if ((tree[deep][i] < sorted[mid]) || (tree[deep][i] == sorted[mid] && same > 0)) {
			flag = 1;
			tree[deep + 1][ls++] = tree[deep][i];
			if (tree[deep][i] == sorted[mid])
				same--;
		} else {
			tree[deep + 1][rs++] = tree[deep][i];
		}
		toleft[deep][i] = toleft[deep][i - 1]+flag;
	}
	build_tree(left, mid, deep + 1);
	build_tree(mid + 1, right, deep + 1);
}
int query(int left, int right, int k, int L, int R, int deep)
{
	if (left == right)
		return tree[deep][left];
    int mid = (L + R) >> 1;
	int x = toleft[deep][left - 1] - toleft[deep][L - 1];//位于left左边的放于左子树中的数字个数
	int y = toleft[deep][right] - toleft[deep][L - 1];//到right为止位于左子树的个数
	int ry = right - L - y;//到right右边为止位于右子树的数字个数
	int cnt = y - x;//[left,right]区间内放到左子树中的个数
	int rx = left - L - x;//left左边放在右子树中的数字个数
	if (cnt >= k) {
		//printf("sss %d %d %d\n", xx++, x, y);
		return query(L + x, L + y - 1, k, L, mid, deep + 1);
	}
	else {
		//printf("qqq %d %d %d\n", xx++, x, y);
		return query(mid + rx + 1, mid + 1 + ry, k - cnt, mid + 1, R, deep + 1);
	}
}
int main()
{
	int m, n;
	int a, b, k;
	int i;
	while (scanf("%d%d", &m, &n) == 2) {
		for (i = 1; i <= m; ++i) {
			scanf("%d", &sorted[i]);
			tree[0][i] = sorted[i];
		}
		sort(sorted + 1, sorted + 1 + m);
		build_tree(1, m, 0);
		for (i = 0; i < n; ++i) {
			scanf("%d%d%d", &a, &b, &k);
			printf("%d\n", query(a, b, k, 1, m, 0));
		}
	}
	return 0;
}