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;
struct interval {
  int start;
  int val;
};
interval V[200000 + 4];
int offset[200000 + 4];
int main() {
  ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  offset[0] = 0;
  int Vi = 0;
  int start = 0, start_v;
  cin >> start_v;
  interval iv;
  for (int i = 1; i < n; i++) {
    int x;
    cin >> x;
    if (x == start_v)
      offset[i] = offset[i - 1];
    else {
      iv.start = start;
      iv.val = start_v;
      V[Vi] = (iv);
      Vi++;
      start = i;
      start_v = x;
      offset[i] = offset[i - 1] + 1;
    }
  }
  iv.start = start;
  iv.val = start_v;
  V[Vi] = (iv);
  for (int i = 0; i < m; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    l--;
    r--;
    if (offset[l] == offset[r] && V[offset[l]].val == x)
      printf("-1\n");
    else {
      if (offset[l] == offset[r])
        printf("%d\n", l + 1);
      else if (V[offset[l]].val == x)
        printf("%d\n", V[offset[l] + 1].start + 1);
      else
        printf("%d\n", l + 1);
    }
  }
  return 0;
}