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;
using ll = long long;
using ld = long double;
const int N = 1e6 + 10;
const ll INF = 1e18 * 3;
pair<int, int> s[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> s[i].first;
    s[i].second = i;
  }
  for (int i = n - 1; i >= 1; i--) {
    if (s[i].first != s[i + 1].first)
      s[i].second = i;
    else
      s[i].second = s[i + 1].second;
  }
  int l, r, x;
  for (int i = 0; i < m; i++) {
    cin >> l >> r >> x;
    if (s[l].first != x)
      cout << l << endl;
    else {
      if (s[l].second >= r)
        cout << -1 << endl;
      else
        cout << s[l].second + 1 << endl;
    }
  }
}