Problem: You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.

Examples

Input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2


Output

2
6
-1
4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct unionfind {
  unionfind* parent;
  int rank, next;
  unionfind* head() {
    if (parent != parent->parent) {
      parent = parent->head();
    }
    return parent;
  }
  void merge(unionfind* sec) {
    unionfind* fir = head();
    sec = sec->head();
    if (fir->rank > sec->rank) {
      sec->parent = fir;
    } else if (fir->rank < sec->rank) {
      fir->parent = sec;
    } else {
      fir->parent = sec;
      sec->rank++;
    }
  }
};
static unionfind un[1000000];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  int val[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &val[i]);
    un[i].parent = &un[i];
    un[i].rank = 0;
    un[i].next = 2 * n;
  }
  for (int i = 0; i < n - 1; i++) {
    if (val[i] == val[i + 1]) {
      un[i].merge(&un[i + 1]);
    } else {
      un[i].head()->next = i + 1;
    }
  }
  for (int i = 0; i < q; i++) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    l--, r--;
    if (val[l] != x) {
      printf("%d", l + 1);
    } else if (un[l].head()->next <= r) {
      printf("%d", un[l].head()->next + 1);
    } else {
      printf("-1");
    }
    printf("\n");
  }
  return 0;
}