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 int N = 2e5 + 10;
int n, m;
int a[N];
int pr[N];
long long sum[N];
stack<pair<int, int> > st;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum[i] = sum[i - 1] + a[i];
  }
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && st.top().first == a[i]) {
      st.pop();
    }
    if (st.empty()) {
      pr[i] = -1;
    } else {
      pr[i] = st.top().second;
    }
    st.push({a[i], i});
  }
  while (m--) {
    int l, r, x;
    cin >> l >> r >> x;
    if (a[r] != x) {
      cout << r << "\n";
    } else {
      if (l <= pr[r]) {
        cout << pr[r] << "\n";
      } else {
        cout << -1 << "\n";
      }
    }
  }
  return 0;
}