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;
using pii = pair<int, int>;
const int N = 2e5;
int32_t n, m, a[N + 1], b[N + 1];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout << fixed;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = n, j = n + 1; i >= 1; i--) {
    b[i] = j;
    if (a[i - 1] != a[i]) j = i;
  }
  while (m-- > 0) {
    int l, r, x;
    cin >> l >> r >> x;
    cout << (a[l] != x ? l : b[l] <= r ? b[l] : -1) << '\n';
  }
  return 0;
}