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:
/*
	The idea here is by using reference table to lookup
	index of next different number
*/

var nm = readline().split(' ').map(function(x){ return parseInt(x) - 1; });
var arr = readline().split(' ');

var arr_jmp = []; //reference table

//fill reference table backward
arr_jmp[nm[0]] = arr.length; 
var current_val = parseInt(arr[nm[0]]);
for(var i = nm[0] - 1; i >= 0; i--)
{
	if(parseInt(arr[i]) == current_val) {
		arr_jmp[i] = arr_jmp[i + 1];
	} else {
		arr_jmp[i] = i + 1;
		current_val = parseInt(arr[i]);
	}
}

// processing query
var result = "";
for(var i = 0; i <= nm[1]; i++)
{
	var params = readline().split(' ');
	result += query(parseInt(params[0]) - 1, parseInt(params[1]) - 1, parseInt(params[2]));
	result += "\n";
}

print(result); //print only once, it will save a lot of time

function query(l, r, x)
{
	if(parseInt(arr[l]) == x){
		if(arr_jmp[l] > r)
			return -1;
		else
			return arr_jmp[l] + 1; // +1 because in problem, index start from 1
	} else {
		return l + 1; // +1 because in problem, index start from 1
	}
}