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 main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int t[n];
  int dd[n];
  scanf("%d", &t[0]);
  dd[0] = 0;
  for (int i = 1; i < n; i++) {
    scanf("%d", &t[i]);
    if (t[i] == t[i - 1]) {
      dd[i] = dd[i - 1] + 1;
    } else
      dd[i] = 0;
  }
  int a, b, c;
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &a, &b, &c);
    a--;
    b--;
    if (t[b] != c) {
      printf("%d", b + 1);
    } else {
      if (dd[b] >= b - a) {
        printf("-1");
      } else
        printf("%d", b - dd[b]);
    }
    printf("\n");
  }
  return 0;
}