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(Number);
var arr = readline().split(' ').map(Number);
var arr_jmp = []; // this is the reference table

for(var i = 0; i < nm[0]; i++) {
	if(arr[i-1] !== arr[i]) 
		arr_jmp[i] = i;
	else 
		arr_jmp[i] = arr_jmp[i-1]
}

var result = "";

for(var i = 0; i < nm[1]; i++)
{
	var lrx = readline().split(' ').map(Number);
	var l = lrx[0]; var r = lrx[1]; var x = lrx[2];

	if (arr[r-1] !== x) {
	    result += r;
	} else if (arr_jmp[r-1] >= l) {
	    result += arr_jmp[r-1];
	} else {
	    result += '-1';
	}
	
	result += '\n';
}

print(result);