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 + 10;
vector<int> pos[maxn];
int num[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    num[i] = a;
    pos[a].push_back(i);
  }
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    l--;
    int t1 = lower_bound(pos[x].begin(), pos[x].end(), l) - pos[x].begin();
    int t2 = lower_bound(pos[x].begin(), pos[x].end(), r) - pos[x].begin();
    if (t2 - t1 == r - l) {
      cout << "-1\n";
      continue;
    }
    if (num[l] != x) {
      cout << l + 1 << "\n";
      continue;
    }
    int lo = l, hi = r + 1;
    while (hi - lo > 1) {
      int mid = (hi + lo) / 2;
      int t = lower_bound(pos[x].begin(), pos[x].end(), mid) - pos[x].begin();
      if (t - t1 != mid - l)
        hi = mid;
      else
        lo = mid;
    }
    cout << hi << "\n";
  }
}