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;
void faster() {
  if (0) {
    freopen(
        "input"
        ".in",
        "r", stdin);
    freopen(
        "input"
        ".out",
        "w", stdout);
  }
}
int n, m, a[200100], l, r, x, t, lf[200100];
int main() {
  faster();
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= int(n); i++) {
    scanf("%d", &a[i]);
    if (i > 1 && a[i] != a[i - 1]) {
      t = i - 1;
      while (t > 0 && a[t] == a[i - 1]) {
        lf[t] = i;
        t--;
      }
    }
  }
  while (m--) {
    scanf("%d %d %d", &l, &r, &x);
    if (a[l] != x) {
      printf("%d\n", l);
      continue;
    }
    if (!lf[l] || lf[l] > r) {
      printf("%d\n", -1);
      continue;
    }
    printf("%d\n", lf[l]);
  }
  return 0;
}