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 JAVA):

Solution:


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C622 {

	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
	static int n, m, l, r, x;
	static StringTokenizer st;
	static int[] a = new int[200200];
	static int[] z = new int[200200];

	public static void main(String[] args) throws IOException {
		st = new StringTokenizer(br.readLine());
		n = Integer.parseInt(st.nextToken());
		m = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(st.nextToken());
		}

		z[0] = -1;

		for (int i = 1; i < n; i++) {
			if (a[i] != a[i - 1])
				z[i] = i - 1;
			else
				z[i] = z[i - 1];
		}

		for (int i = 0; i < m; i++) {
			st = new StringTokenizer(br.readLine());
			l = Integer.parseInt(st.nextToken());
			r = Integer.parseInt(st.nextToken());
			x = Integer.parseInt(st.nextToken());

			if (a[r - 1] != x) {
				pr.println(r);
			} else {
				if (z[r - 1] >= l - 1) {
					pr.println(z[r - 1] + 1);
				} else {
					pr.println(-1);
				}
			}
		}

		br.close();
		pr.flush();
		pr.close();
	}
}