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;
bool deb = false;
int i;
const int N = 200011;
int a[N], ans[N];
void pre(int n) {
  int i;
  stack<int> s;
  s.push(1);
  for (i = 2; i < n + 1; ++i) {
    if (a[s.top()] == a[i]) {
      s.push(i);
    } else {
      while (!s.empty() && a[s.top()] != a[i]) {
        ans[s.top()] = i;
        s.pop();
      }
      s.push(i);
    }
  }
  while (!s.empty()) ans[s.top()] = 1000000007, s.pop();
}
int main() {
  int i, j, n, q;
  scanf("%d", &n), scanf("%d", &q);
  for (i = 1; i < n + 1; ++i) scanf("%d", &a[i]);
  pre(n);
  while (q--) {
    int l, r, x;
    scanf("%d", &l), scanf("%d", &r), scanf("%d", &x);
    if (a[l] != x) {
      printf("%d\n", l);
    } else {
      int tmp = ans[l];
      if (tmp >= l && tmp <= r)
        printf("%d\n", tmp);
      else
        printf("%d\n", -1);
    }
  }
  return 0;
}