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 MAXN = 1e6 + 5;
int n, m;
int a[MAXN], p1[MAXN], p2[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = n - 1; i >= 0; i--) {
    if (i == (n - 1))
      p2[i] = -1;
    else if (a[i + 1] != a[i])
      p2[i] = i + 1;
    else
      p2[i] = p2[i + 1];
  }
  while (m) {
    m--;
    int l, r, x;
    cin >> l >> r >> x;
    if (a[l - 1] != x)
      cout << l << endl;
    else if (p2[l - 1] <= (r - 1) && p2[l - 1] != -1)
      cout << p2[l - 1] + 1 << endl;
    else
      cout << -1 << endl;
  }
  return 0;
}