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 int MAXN = 2 * 1e5 + 100;
int niz[MAXN], inter[MAXN], n, m, a, l, r, x;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &niz[i]);
  }
  inter[n - 1] = n - 1;
  for (int i = n - 2; i >= 0; i--) {
    if (niz[i] == niz[i + 1])
      inter[i] = inter[i + 1];
    else
      inter[i] = i;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &l, &r, &x);
    l--;
    r--;
    if (niz[l] == x) {
      l = inter[l];
      l++;
      if (l > r)
        printf("-1\n");
      else
        printf("%d\n", l + 1);
    } else {
      printf("%d\n", l + 1);
    }
  }
  return 0;
}