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 GCC optimize("O3")
#pragma GCC target("sse4")
const double PI = acos(-1.0);
const double EPS = 1e-9;
const long long mod = 1000000007;
const long long INF = 1e9;
const long long LINF = 1e9;
const long long MAX = 2e5 + 7;
const long long MAXN = 1e6 + 1;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  vector<int> fake_next(n, -1);
  int curr = 0;
  vector<int> t;
  while (curr < n - 1) {
    int now = curr;
    if (a[curr] != a[curr + 1]) {
      fake_next[curr] = curr + 1;
      ++curr;
      continue;
    }
    t.clear();
    t.push_back(curr);
    ++curr;
    while (curr < n && a[curr] == a[curr - 1]) t.push_back(curr++);
    if (curr != n) {
      for (auto i : t) fake_next[i] = curr;
    }
  }
  int l, r, x;
  for (int i = (0); i < (m); ++i) {
    cin >> l >> r >> x;
    --l, --r;
    if (a[l] != x) {
      cout << l + 1 << endl;
      continue;
    }
    if (fake_next[l] == -1 || fake_next[l] > r)
      cout << -1 << endl;
    else
      cout << fake_next[l] + 1 << endl;
  }
  return 0;
}