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 v[200005];
int nxt[200005];
vector<int> idx[1000005];
int main() {
  int n, m, x, l, r, first, ans;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    idx[v[i]].push_back(i);
  }
  nxt[n - 1] = n;
  for (int i = n - 1 - 1; i >= 0; i--) {
    if (v[i] == v[i + 1])
      nxt[i] = nxt[i + 1];
    else
      nxt[i] = i + 1;
  }
  while (m--) {
    scanf("%d %d %d", &l, &r, &x);
    l -= 1;
    r -= 1;
    vector<int>::iterator i = lower_bound(idx[x].begin(), idx[x].end(), l);
    if (i == idx[x].end()) {
      ans = l + 1;
    } else {
      first = *i;
      if (l < first) {
        ans = l + 1;
      } else if (nxt[l] <= r) {
        ans = nxt[l] + 1;
      } else {
        ans = -1;
      }
    }
    printf("%d\n", ans);
  }
}