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>
const int pi = acos(-1);
using namespace std;
const int M = 2 * 100005;
int n, q, a[M], par[M], l, r, x;
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) par[i] = i;
  for (int i = 2; i <= n; i++)
    if (a[i] == a[i - 1]) par[i] = par[i - 1];
  while (q--) {
    scanf("%d %d %d", &l, &r, &x);
    if (par[l] == par[r]) {
      if (a[l] == x)
        printf("-1\n");
      else
        printf("%d\n", l);
      continue;
    }
    if (x != a[l]) {
      printf("%d\n", l);
      continue;
    }
    if (x != a[r]) {
      printf("%d\n", r);
      continue;
    }
    printf("%d\n", par[r] - 1);
  }
  return 0;
}