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[200001], rez[200001], p[200001];
int main() {
  int i, n, m, pretpoz, trenpoz, pretelement, l, r, x;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  p[1] = 0;
  trenpoz = 0;
  pretpoz = 0;
  pretelement = a[1];
  for (i = 2; i <= n; i++) {
    if (a[i] == a[i - 1])
      p[i] = p[i - 1];
    else
      p[i] = i - 1;
  }
  for (i = 0; i < m; i++) {
    scanf("%d%d%d", &l, &r, &x);
    if (p[r] >= l && a[p[r]] != x)
      rez[i] = p[r];
    else {
      if (a[r] != x)
        rez[i] = r;
      else
        rez[i] = -1;
    }
  }
  for (i = 0; i < m; i++) printf("%d\n", rez[i]);
  return 0;
}