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 long long N = 2e5 + 7;
const string alpha = "abcdefghijklmnopqrstuvwxyz";
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T powm(T a, T b, long long m) {
  T cnt = 1;
  while (b > 0) {
    if (b % 2 == 1) cnt = (cnt * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return cnt % m;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long sqr(long long x) { return x * x; }
void solve() {
  int n, m, l, r, x;
  cin >> n >> m;
  vector<int> a(n + 1), cnt(n + 1);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = n - 1; i > 0; --i) {
    if (a[i] == a[i + 1]) cnt[i] = cnt[i + 1] + 1;
  }
  while (m--) {
    cin >> l >> r >> x;
    if (a[l] != x)
      cout << l;
    else if (l + cnt[l] < r)
      cout << l + cnt[l] + 1;
    else
      cout << -1;
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc;
  tc = 1;
  while (tc--) solve();
  return 0;
}