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.StreamTokenizer;
import java.util.Scanner;

/**
 * Created by mmaikovych on 10/02/2016.
 */
public class CF_C {
    public static void main(String[] args) throws IOException {
        StreamTokenizer input  = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        input.nextToken();
        int n = (int)input.nval;
        input.nextToken();
        int m = (int)input.nval;
        int pointers[] = new int[n];
        int numbers[] = new int[n];
        int numbersIndexes[] = new int[n];
        int totalumbersCount = 1;
        pointers[0] = totalumbersCount - 1;
        input.nextToken();
        numbers[0] = (int)input.nval;
        int number;
        for (int i = 1; i < n; i++) {
            input.nextToken();
            number = (int)input.nval;
            if (number != numbers[totalumbersCount - 1]) {
                numbers[totalumbersCount] = number;
                numbersIndexes[totalumbersCount] = i;
                totalumbersCount++;
            }
            pointers[i] = totalumbersCount - 1;
        }
        StringBuilder result = new StringBuilder(m*10);
        int l, r, x;
        int pointer;
        for (int i = 0; i < m; i++) {
            input.nextToken();
            l = (int)input.nval;
            input.nextToken();
            r = (int)input.nval;
            input.nextToken();
            x = (int)input.nval;
            pointer = pointers[l - 1];
            while ((pointer < totalumbersCount) && (numbers[pointer] == x) && (numbersIndexes[pointer] < r)) {
                pointer++;
            }
            if ((pointer >= totalumbersCount)||(numbers[pointer] == x)||(numbersIndexes[pointer] >= r)) {
                result.append(-1);
                result.append("\n");
            } else {
                result.append(Math.max(numbersIndexes[pointer] + 1, l));
                result.append("\n");
            }
        }
        System.out.println(result.toString());
    }
}