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;
struct edge {
  long long from, to, w;
  edge(long long from, long long to, long long w) : from(from), to(to), w(w) {}
  bool operator<(const edge& e) const { return ((w > e.w)); }
};
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long fx(long long n) {
  return ((n % 1000000007) + 1000000007) % 1000000007;
}
void O_o() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
int main() {
  O_o();
  int n, m, a, b, x;
  cin >> n >> m;
  int arr[n + 1], sum[n + 5];
  sum[0] = arr[0] = 0;
  for (int i = 1; i <= int(n); i++) {
    cin >> arr[i];
    sum[i] = (arr[i] == arr[i - 1]) ? sum[i - 1] : i;
  }
  while (m--) {
    cin >> a >> b >> x;
    if (arr[sum[b]] != x)
      cout << b;
    else if (sum[b] > a)
      cout << sum[b] - 1;
    else
      cout << -1;
    cout << endl;
  }
  return 0;
}