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 nxt[200005];
vector<int> v;
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int aux;
    cin >> aux;
    v.push_back(aux);
  }
  memset(nxt, -1, sizeof nxt);
  for (int i = n - 2; i >= 0; i--) {
    if (v[i] == v[i + 1]) {
      nxt[i] = nxt[i + 1];
    } else {
      nxt[i] = i + 1;
    }
  }
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    if (v[a] != c) {
      cout << a + 1 << "\n";
    } else {
      if (nxt[a] == -1 or nxt[a] > b) {
        cout << -1 << "\n";
      } else {
        cout << nxt[a] + 1 << "\n";
      }
    }
  }
  return 0;
}