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 = 204800;
int a[maxn], b[maxn];
int main() {
  int n, q;
  int l, r, x;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  b[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] != a[i - 1])
      b[i] = i;
    else
      b[i] = b[i - 1];
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d", &l, &r, &x);
    if (l == r) {
      if (a[l] == x)
        printf("-1\n");
      else
        printf("%d\n", l);
    } else if (x != a[r])
      printf("%d\n", r);
    else if (b[l] == b[r])
      printf("-1\n");
    else {
      printf("%d\n", lower_bound(b + l, b + r + 1, b[r]) - b - 1);
    }
  }
}