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[10000000];
int l[10000000];
int r[10000000];
int x[10000000];
int z[10000000];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &l[i], &r[i], &x[i]);
    l[i]--;
    r[i]--;
  }
  z[0] = -1;
  for (int i = 1; i < n; i++) {
    if (a[i - 1] != a[i])
      z[i] = i - 1;
    else
      z[i] = z[i - 1];
  }
  for (int i = 0; i < m; i++) {
    if (a[r[i]] != x[i])
      printf("%d\n", r[i] + 1);
    else if (z[r[i]] >= l[i])
      printf("%d\n", z[r[i]] + 1);
    else
      printf("%d\n", -1);
  }
  return 0;
}