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[200005];
int n;
int cnt[200005];
int main() {
  int m;
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 0; i <= (n - 1); i++) scanf("%d", &a[i]);
  int i = 0;
  while (i < n) {
    int j = i + 1;
    while (a[j] == a[i]) j++;
    cnt[i] = j - i;
    i = j;
  }
  for (int i = 1; i <= (n - 1); i++)
    if (cnt[i] == 0) cnt[i] = cnt[i - 1] - 1;
  for (int i = 0; i <= (m - 1); i++) {
    int a1, b1, c1;
    scanf("%d", &a1);
    scanf("%d", &b1);
    scanf("%d", &c1);
    int ans = -1;
    int in = a1 - 1;
    while (in < b1) {
      if (a[in] != c1) {
        ans = in + 1;
        break;
      }
      in += cnt[in];
    }
    cout << ans << endl;
  }
}