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;
int n, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  long long int i;
  vector<int> v(n + 1, 0);
  for (i = 1; i <= n; i++) {
    cin >> v[i];
  }
  int dp[n + 1], l[m + 1], r[m + 1];
  memset(dp, 0, sizeof(dp));
  dp[n] = n;
  for (long long int i = n - 1; i >= 1; i--) {
    if (v[i] == v[i + 1])
      dp[i] = dp[i + 1];
    else
      dp[i] = i;
  }
  for (i = 1; i <= m; i++) {
    int x;
    cin >> l[i] >> r[i] >> x;
    if (v[l[i]] != x)
      cout << l[i] << "\n";
    else if (dp[l[i]] >= r[i])
      cout << -1 << "\n";
    else
      cout << dp[l[i]] + 1 << "\n";
  }
  return 0;
}