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 long long M = 1e6 + 10, MOD = 1e9 + 7;
vector<int> vec[M], ans[M];
int mx;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  map<int, set<int>> mp;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    vec[a].push_back(i);
    mp[a].insert(i);
    mx = max(mx, a);
  }
  for (int i = 1; i <= mx; i++) {
    if (vec[i].size()) {
      int target = 0;
      for (int j = 0; j < vec[i].size(); j++) {
        if (j == 0) {
          if (vec[i][j] == 1)
            ans[i].push_back(target);
          else {
            ans[i].push_back(vec[i][j] - 1);
            target = vec[i][j] - 1;
          }
        } else {
          if (vec[i][j] - vec[i][j - 1] > 1) {
            ans[i].push_back(vec[i][j] - 1);
            target = vec[i][j] - 1;
          } else {
            ans[i].push_back(target);
          }
        }
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    int l, r, val;
    cin >> l >> r >> val;
    if (mp[val].find(l) == mp[val].end()) {
      cout << l << '\n';
      continue;
    }
    if (mp[val].find(r) == mp[val].end()) {
      cout << r << '\n';
      continue;
    }
    int L = 0, R = vec[val].size() - 1;
    int sol;
    while (L <= R) {
      int mid = L + ((R - L) >> 1);
      if (vec[val][mid] == r) {
        sol = ans[val][mid];
        break;
      } else if (vec[val][mid] < r)
        L = mid + 1;
      else
        R = mid - 1;
    }
    cout << (sol >= l ? sol : -1);
    cout << "\n";
  }
  return 0;
}