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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 42;
int dist[N] = {0}, a[N] = {0};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n = 0, m = 0;
  cin >> n >> m;
  int cur = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (!i) {
      dist[i] = cur = 42;
      continue;
    }
    if (a[i] == a[i - 1])
      cur++;
    else
      cur = 1;
    dist[i] = cur;
  }
  while (m--) {
    int l, r, x;
    cin >> l >> r >> x;
    --l;
    --r;
    if (a[r] != x) {
      cout << r + 1 << endl;
      continue;
    }
    if (l <= r - dist[r]) {
      cout << r - dist[r] + 1 << endl;
      continue;
    }
    cout << -1 << endl;
  }
  return 0;
}