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.IOException;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Scanner;

public class CodeC {

	/**
	 * @param args
	 * @throws IOException 
	 */
	public static void main(String[] args) throws IOException {
		StreamTokenizer in = new StreamTokenizer(System.in);
		PrintWriter out = new PrintWriter(System.out);
		in.nextToken();
		int n = (int) in.nval;
		in.nextToken();
		int m = (int) in.nval;
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			in.nextToken();
			arr[i] = (int) in.nval;
		}
		int[] help = new int[n];
		for (int i = 1; i < n; i++) {
			if (arr[i - 1] == arr[i]) {
				help[i] = help[i - 1] + 1;
			}
		}
		int l = 0;
		int r = 0;
		int x = 0;
		for (int i = 0; i < m; i++) {
			in.nextToken();
			l = (int) in.nval - 1;
			in.nextToken();
			r = (int) in.nval - 1;
			in.nextToken();
			x = (int) in.nval;
			if (l == r) {
				if (arr[l] == x) {
					out.println(-1);
				} else {
					out.println(l + 1);
				}
			} else {
				while (r > l && arr[r] == x) {
					if (help[r] > 0) {
						r -= help[r] + 1;
					} else {
						r--;
					}
				}
				if (r < l) {
					out.println(-1);
				} else {
					out.println(r + 1);
				}
			}
		}
		out.flush();
		out.close();
	}
}