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;
const int N = (int)2e5 + 5;
int seg[N], ans[N];
vector<pair<int, int> > beg[N], fin[N];
set<pair<int, int> > pending;
set<pair<int, int> >::iterator it1, it2;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> seg[i];
  for (int q = 0; q < m; q++) {
    int l, r, x;
    cin >> l >> r >> x;
    beg[l].emplace_back(x, q);
    fin[r].emplace_back(x, q);
  }
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> xq : beg[i]) pending.insert(xq);
    it2 = pending.lower_bound(make_pair(seg[i], 0));
    it1 = pending.begin();
    while (it1 != it2) ans[(it1++)->second] = i;
    pending.erase(pending.begin(), it2);
    it1 = it2 = pending.lower_bound(make_pair(seg[i] + 1, 0));
    while (it1 != pending.end()) ans[(it1++)->second] = i;
    pending.erase(it2, pending.end());
    for (pair<int, int> xq : fin[i])
      if ((it1 = pending.find(xq)) != pending.end()) {
        ans[xq.second] = -1;
        pending.erase(it1);
      }
  }
  for (int q = 0; q < m; q++) cout << ans[q] << '\n';
}