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::sync_with_stdio(false);
  cin.tie(0);
  int n, m, l, r, input;
  cin >> n >> m;
  vector<int> preprocess(n + 1), arr(n + 1);
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  preprocess[1] = 0;
  for (int i = 2; i <= n; ++i) {
    if (arr[i] == arr[i - 1])
      preprocess[i] = preprocess[i - 1];
    else
      preprocess[i] = i - 1;
  }
  while (m--) {
    cin >> l >> r >> input;
    if (arr[r] == input && preprocess[r] < l)
      cout << "-1\n";
    else {
      if (arr[r] != input)
        cout << r << "\n";
      else
        cout << preprocess[r] << "\n";
    }
  }
}