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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n + 1), b(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = n, j = n + 1; i >= 1; --i) {
    if (i != n && a[i] != a[i + 1]) {
      j = i + 1;
    }
    b[i] = j;
  }
  for (int i = 0; i < m; ++i) {
    long long l, r, x;
    cin >> l >> r >> x;
    long long ind = l;
    bool flag = false;
    if (a[ind] == x) {
      ind = b[ind];
      if (ind <= r) {
        cout << ind << endl;
      } else {
        cout << -1 << endl;
      }
    } else {
      cout << ind << endl;
    }
  }
  return 0;
}