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 + 10, md = 1e9 + 7;
const double Eps = 0.01;
using namespace std;
int n, a[Maxn], nex[Maxn], m, l, r, x;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[n + 1] = -1;
  for (int i = 1; i <= n; i++)
    if (a[i] != a[i + 1])
      nex[i] = i + 1;
    else {
      if (a[i - 1] == a[i])
        nex[i] = nex[i - 1];
      else
        for (int j = i + 2; j <= n + 1; j++)
          if (a[j] != a[i]) {
            nex[i] = j;
            break;
          }
    }
  while (m--) {
    scanf("%d %d %d", &l, &r, &x);
    if (a[l] != x)
      printf("%d\n", l);
    else if (nex[l] < r + 1)
      printf("%d\n", nex[l]);
    else
      puts("-1");
  }
}