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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
void __print(long long first) { cerr << first; }
void __print(long first) { cerr << first; }
void __print(unsigned first) { cerr << first; }
void __print(unsigned long first) { cerr << first; }
void __print(unsigned long long first) { cerr << first; }
void __print(float first) { cerr << first; }
void __print(double first) { cerr << first; }
void __print(long double first) { cerr << first; }
void __print(char first) { cerr << '\'' << first << '\''; }
void __print(const char *first) { cerr << '\"' << first << '\"'; }
void __print(const string &first) { cerr << '\"' << first << '\"'; }
void __print(bool first) { cerr << (first ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &first) {
  cerr << '{';
  __print(first.first);
  cerr << ',';
  __print(first.second);
  cerr << '}';
}
template <typename T>
void __print(const T &first) {
  long long f = 0;
  cerr << '{';
  for (auto &i : first) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, 0};
const long long N = 1e6 + 5;
long long n, m;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> z(n);
  z[n - 1] = n + 1;
  for (long long i = n - 2; i >= 0; i--) {
    if (a[i] == a[i + 1])
      z[i] = z[i + 1];
    else
      z[i] = i + 1;
  }
  for (long long j = 0; j < m; j++) {
    long long l, r, first;
    cin >> l >> r >> first;
    l--;
    r--;
    long long ans = -1;
    if (a[l] != first)
      ans = l + 1;
    else {
      long long d = z[l];
      if (d <= r) ans = d + 1;
    }
    cout << ans << "\n";
  }
  return 0;
}