#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 100010, M = 50010;

//快读
int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
int n, m;

struct A {
    int id, x;
    const bool operator<(const A &t) const {
        return x < t.x;
    }
} a[N];

struct Q {
    int id;      //第几号查询
    int l, r, k; //左右区间，第k小
} q[M];

//树状数组
int tr[N];
//查询x的个数和
int query(int x) {
    int sum = 0;
    for (; x; x -= x & -x) sum += tr[x];
    return sum;
}
// x位置上增加c
void update(int x, int c) {
    for (; x <= n; x += x & -x) tr[x] += c;
}

//每个询问的答案数组
int ans[M];

//[l,r]:排序后数组上的区间
//[x,y]:询问集合的区间
void solve(int l, int r, int x, int y) {
    if (x > y) return; //查询的子区间非法，返回

    if (l == r) {
        for (int i = x; i <= y; i++) ans[q[i].id] = a[l].x;
        return;
    }

    int mid = (l + r) >> 1;
    for (int i = l; i <= mid; i++) update(a[i].id, 1);

    int i = x, j = y, t;

    while (i <= j) {
        while (i <= j && q[i].k <= query(q[i].r) - query(q[i].l - 1)) i++;

        while (i <= j && q[j].k > (t = query(q[j].r) - query(q[j].l - 1))) {
            q[j].k -= t;
            j--;
        }

        if (i < j)
            swap(q[i], q[j]);
    }

    for (int i = l; i <= mid; i++) update(a[i].id, -1);

    solve(l, mid, x, j), solve(mid + 1, r, i, y);
}

int main() {
    n = read(), m = read();

    for (int i = 1; i <= n; i++) a[i].x = read(), a[i].id = i;

    //查询请求
    for (int i = 1; i <= m; i++) {
        q[i].l = read(), q[i].r = read(), q[i].k = read();
        q[i].id = i;
    }

    sort(a + 1, a + 1 + n); //将原数组的二元组按值大小进行由小到大的排序，同步记录了每个数值对应的序号

    //离线处理
    //二分区间为[1,n] ,查询区间q[1]~q[m]???
    solve(1, n, 1, m);

    //输出每次询问的答案
    for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);

    return 0;
}