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
#include <bits/stdc++.h>
using namespace std;
int N, M, a[200010], dp[200010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  int i, l, r, x;
  for (i = 1; i <= N; i++) cin >> a[i];
  for (i = N; i > 0; i--) {
    dp[i] = i;
    if (a[i + 1] == a[i]) dp[i] = dp[i + 1];
  }
  while (M--) {
    cin >> l >> r >> x;
    if (a[l] != x)
      cout << l << "\n";
    else {
      if (dp[l] >= r)
        cout << -1 << "\n";
      else
        cout << dp[l] + 1 << "\n";
    }
  }
  return 0;
}
