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 n, m, nn[200005], mm[200005], t1, t2, t3;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie();
  cin >> n >> m;
  memset(mm, -1, sizeof mm);
  for (int i = 0; i < n; i++) {
    cin >> nn[i];
    if (i)
      if (nn[i] != nn[i - 1])
        mm[i] = i - 1;
      else
        mm[i] = mm[i - 1];
  }
  for (int i = 0; i < m && cin >> t1 >> t2 >> t3; i++)
    if (nn[t2 - 1] != t3)
      cout << t2 << "\n";
    else if (mm[t2 - 1] + 1 >= t1)
      cout << mm[t2 - 1] + 1 << "\n";
    else
      cout << "-1\n";
  return 0;
}