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;
priority_queue<pair<int, int> > qmin, qmax;
vector<int> op[200001], cl[200001];
int n, m, tab[200001], x, y, v[200001], ans[200001];
bool met[200001];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int a = 1; a <= n; a++) cin >> tab[a];
  for (int a = 1; a <= m; a++) {
    cin >> x >> y >> v[a];
    op[x].push_back(a);
    cl[y].push_back(a);
  }
  for (int a = 1; a <= n; a++) {
    for (int b = 0; b < op[a].size(); b++) {
      qmax.push(make_pair(v[op[a][b]], op[a][b]));
      qmin.push(make_pair(-v[op[a][b]], op[a][b]));
    }
    while (1) {
      if (qmax.empty()) break;
      if (met[(qmax.top()).second] == 1)
        qmax.pop();
      else if ((qmax.top()).first != tab[a]) {
        met[(qmax.top()).second] = 1;
        ans[(qmax.top()).second] = a;
        qmax.pop();
      } else
        break;
    }
    while (1) {
      if (qmin.empty()) break;
      if (met[(qmin.top()).second] == 1)
        qmin.pop();
      else if (-(qmin.top()).first != tab[a]) {
        met[(qmin.top()).second] = 1;
        ans[(qmin.top()).second] = a;
        qmin.pop();
      } else
        break;
    }
    for (int b = 0; b < cl[a].size(); b++) {
      if (!met[cl[a][b]]) {
        ans[cl[a][b]] = -1;
        met[cl[a][b]] = 1;
      }
    }
  }
  for (int a = 1; a <= m; a++) cout << ans[a] << "\n";
  return 0;
}