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.*;
import java.util.*;
import java.math.*;

public class C622 {

	public static BufferedReader read = null;
	public static PrintWriter out = null;
	public static StringTokenizer token = null;
	
	
	public static void solve()
	{
		int n = nint(), m = nint();
		int[] a = narr(n);
		int[] nd = new int[n]; //nd[i] stores the minimum positive value s such that nd[s] != nd[i]
		nd[n-1] = -1;
		for(int i=n-2; i>=0; i--)
		{
			if(a[i] == a[i+1]) nd[i] = nd[i+1];
			else nd[i] = i+1;
		}
		
		for(int i=0; i<m; i++)
		{
			int l = nint()-1, r = nint()-1, x = nint();
			int loc = l;
			if(a[loc] == x)
			{
				loc = nd[loc];
				if(loc != -1 && loc <= r)
				{
					out.println(loc+1);
				}
				else
				{
					out.println("-1");
				}
			}
			else
			{
				out.println(loc+1);
			}
		}
		
	}
	
	public static void main(String[] args)
	{
		read = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.flush();
		out.close();
		
	}
	
	// i/o functions
	public static String next() // returns the next string
	{
		while(token == null || !token.hasMoreTokens())
		{
			try {
				token = new StringTokenizer(read.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}
		return token.nextToken();
	}
	
	public static int nint()
	{
		return Integer.parseInt(next());
	}
	public static long nlong()
	{
		return Long.parseLong(next());
	}
	public static double ndouble()
	{
		return Double.parseDouble(next());
	}
	public static int[] narr(int n)
	{
		int[] a = new int[n];
		for(int i=0; i<n; i++) a[i] = nint();
		return a;
	}
	
	public static long[] nal(int n)
	{
		long[] a = new long[n];
		for(int i=0; i<n; i++) a[i] = nlong();
		return a;
	}
	
	
}