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;
const double PI = 4.0 * atan(1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f;
long long l[200005] = {0}, n[200005];
int main() {
  long long t, s, mark = 0;
  scanf("%lld%lld", &t, &s);
  for (int i = 0; i <= t; i++) {
    if (i != t) scanf("%lld", &n[i]);
    if (i == 0)
      l[i] = 1;
    else if (n[i] != n[i - 1]) {
      for (int j = mark; j <= i; j++) l[j] = i - j;
      mark = i;
    } else if (i == t)
      for (int j = mark; j <= i; j++) l[j] = i - j + 1;
  }
  long long a, b, f, result;
  while (s--) {
    scanf("%lld%lld%lld", &a, &b, &f);
    result = -1;
    for (int i = a - 1; i <= b - 1; i += l[i]) {
      if (n[i] != f) {
        result = i + 1;
        break;
      }
    }
    printf("%lld\n", result);
  }
  return 0;
}