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 a[1000000], b[1000000];
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) b[i] = (a[i] == a[i - 1]) ? b[i - 1] : (i - 1);
  while (m--) {
    int x, y, v;
    scanf("%d%d%d", &x, &y, &v);
    if (a[y] != v)
      printf("%d\n", y);
    else if (b[y] >= x)
      printf("%d\n", b[y]);
    else
      puts("-1");
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}