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;
int main() {
  long n, q, i, x, y, z;
  scanf("%ld", &n);
  scanf("%ld", &q);
  long A[n];
  set<long> bst_pos;
  bst_pos.insert(0);
  for (i = 0; i < n; i++) {
    long x;
    cin >> x;
    A[i] = x;
    if (i == 0) continue;
    if (A[i] != A[i - 1]) bst_pos.insert(i);
  }
  set<long>::iterator it;
  for (i = 0; i < q; i++) {
    scanf("%ld", &x);
    scanf("%ld", &y);
    scanf("%ld", &z);
    x--, y--;
    if (A[x] != z) {
      printf("%ld\n", x + 1);
    } else {
      it = bst_pos.lower_bound(x + 1);
      if (*it <= y && A[*it] != z)
        printf("%ld\n", *it + 1);
      else
        printf("-1\n");
    }
  }
  return 0;
}