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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  int arr[200005];
  int next[200005];
  for (int i = 1; i <= n; i++) {
    next[i] = 200006;
  }
  cin >> arr[1];
  int holding = 1;
  for (int i = 2; i <= n; i++) {
    cin >> arr[i];
    if (arr[i] != arr[holding]) {
      for (int j = holding; j < i; j++) {
        next[j] = i;
      }
      holding = i;
    }
  }
  int l, r, x;
  for (int i = 0; i < m; i++) {
    cin >> l >> r >> x;
    if (arr[l] == x) {
      if (next[l] <= r) {
        cout << next[l] << endl;
      } else
        cout << -1 << endl;
    } else
      cout << l << endl;
  }
}