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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  vector<pair<long long, long long> > p(n, {-1, -1});
  for (long long i = 0; i < n; i++) cin >> v[i];
  for (long long i = 0; i < n; i++) {
    long long c = v[i], j = i;
    while (++j && j < n && v[j] == c)
      ;
    if (j < n) {
      p[i].first = v[j];
      p[i].second = j;
    } else {
      p[i].first = v[i];
      p[i].second = n - 1;
    }
    i = j - 1;
  }
  for (long long i = 0; i < n; i++) {
    long long ele = p[i].first, ind = p[i].second, j = i;
    while (++j && j < n && p[j].first == -1) {
      p[j].first = ele;
      p[j].second = ind;
    }
    i = j - 1;
  }
  while (k--) {
    long long l, r, x;
    cin >> l >> r >> x;
    l--;
    r--;
    if (v[l] != x)
      cout << l + 1 << "\n";
    else if (p[l].first != x && p[l].second <= r)
      cout << (p[l].second) + 1 << "\n";
    else
      cout << "-1"
           << "\n";
  }
}