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 double PI = acos(-1.0);
long long int pos[200005];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  };
  long long int n, m;
  cin >> n >> m;
  long long int ar[n + 2];
  for (long long int i = 1; i <= n; i++) cin >> ar[i];
  pos[1] = -1;
  for (long long int i = 2; i <= n; i++) {
    if (ar[i] == ar[i - 1])
      pos[i] = pos[i - 1];
    else
      pos[i] = i - 1;
  }
  while (m--) {
    long long int l, r, x;
    cin >> l >> r >> x;
    if (ar[r] != x)
      cout << r << '\n';
    else if (pos[r] >= l)
      cout << pos[r] << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}