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 int N = 2e5 + 5;
int n, m, a[N], lg[N];
pair<int, int> mn[N][26], mx[N][26];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < N; i++) {
    lg[i] = lg[i / 2] + 1;
  }
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    mn[i][0] = make_pair(a[i], i);
    mx[i][0] = make_pair(a[i], i);
  }
  for (int i = 1; i < 26; i++) {
    for (int j = 0; j < n; j++) {
      if (j + (1 << i) > n) continue;
      int go = 1 << (i - 1);
      mn[j][i] = min(mn[j][i - 1], mn[j + go][i - 1]);
      mx[j][i] = max(mx[j][i - 1], mx[j + go][i - 1]);
    }
  }
  for (int i = 0; i < m; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    l--;
    r--;
    int ll = lg[r - l + 1];
    int go = (1 << ll);
    if (x == min(mn[l][ll].first, mn[r - go + 1][ll].first) &&
        min(mn[l][ll].first, mn[r - go + 1][ll].first) ==
            max(mx[l][ll].first, mx[r - go + 1][ll].first)) {
      cout << -1 << "\n";
    } else {
      pair<int, int> val1 = min(mn[l][ll], mn[r - go + 1][ll]);
      pair<int, int> val2 = max(mx[l][ll], mx[r - go + 1][ll]);
      if (val1.first != x) {
        assert(a[val1.second] != x);
        assert(val1.second >= l && val1.second <= r);
        cout << val1.second + 1 << "\n";
      } else {
        assert(a[val2.second] != x);
        assert(val2.second >= l && val2.second <= r);
        cout << val2.second + 1 << "\n";
      }
    }
  }
  return 0;
}