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 a[10000000], b[10000000], memo[10000000], pos[10000000];
int min(int l, int r) {
  if (l <= r) {
    return l;
  }
  return r;
}
int main() {
  int n, m, i, j, k, ind, p, l, x;
  fscanf(stdin, "%d %d", &n, &m);
  for (i = 1; i <= n; i++) {
    fscanf(stdin, "%d", &a[i]);
  }
  i = 1;
  ind = 1;
  while (i <= n) {
    j = i;
    k = a[i];
    memo[ind] = k;
    pos[ind] = i;
    while (j <= n && a[j] == k) {
      b[j] = ind;
      j++;
    }
    i = j;
    ind++;
  }
  for (i = 1; i <= m; i++) {
    fscanf(stdin, "%d %d %d", &p, &l, &x);
    if (b[p] == b[l]) {
      if (x == a[p]) {
        cout << "-1" << endl;
      } else {
        cout << p << endl;
      }
    } else {
      ind = b[p];
      if (memo[b[p]] != x) {
        cout << p << endl;
      } else {
        ind = 1;
        while (pos[b[p] + ind] == x) {
          ind++;
        }
        cout << pos[b[p] + ind] << endl;
      }
    }
  }
  return 0;
}