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>
int main() {
  int k, i, j, n, m, a[200001], pos[200001], l, r, value;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) scanf("%d", a + i);
  i = 0;
  while (i < n) {
    j = i;
    while (j < n && a[j] == a[i]) j++;
    if (j < n)
      for (k = i; k < j; k++) pos[k] = j;
    i = j;
  }
  for (i = 0; i < m; i++) {
    scanf("%d%d%d", &l, &r, &value);
    if (a[l - 1] != value)
      printf("%d\n", l);
    else if (pos[l - 1] >= l - 1 && pos[l - 1] <= r - 1 &&
             a[pos[l - 1]] != value)
      printf("%d\n", pos[l - 1] + 1);
    else
      printf("-1\n");
  }
  return 0;
}