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.PrintWriter;


public class C622 {

	/**
	 * @param args
	 */
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] line = br.readLine().split(" ");
		int n = Integer.parseInt(line[0]);
		long m = Long.parseLong(line[1]);
		line = br.readLine().split(" ");
		int[] a = new int[n+1];
		int[] nextDif = new int[n+1];
		for (int i =1; i<=n; i++) {
			a[i] = Integer.parseInt(line[i-1]);
			if (i == 1) {
				continue;
			}
			int base = a[i-1];
			if (base == a[i]) {
				continue;
			}
			for (int j =i-1 ; j>=0 && a[j] == base;j--) {
				nextDif[j] = i;
			}
		}
		int base = a[n];
		for (int j =n ; j>=0 && a[j] == base;j--) {
			nextDif[j] = Integer.MAX_VALUE;
		}
		PrintWriter pw = new PrintWriter(System.out);
		for (int k = 0; k < m; k++) {
			line = br.readLine().split(" ");
			int l = Integer.parseInt(line[0]);
			int r = Integer.parseInt(line[1]);
			long val = Long.parseLong(line[2]);
			if (a[l] != val) {
				pw.format("%d\n", l);
			} else if (nextDif[l] <= r) {
				pw.format("%d\n", nextDif[l]);
 			} else {
 				pw.format("-1\n");
 			}
		}
		pw.flush();
		pw.close();
		br.close();

	}

}