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 PYTHON):

Solution:
##########################
## Codeforces contest 
##########################
## Eric Breno B. - UFCG - BRAZIL
##########################
	
m, n = map(int, raw_input().split())
eles = [0] * 200005
fodase = [0] *200005
cu = map(int, raw_input().split())
for i in xrange(1, len(cu)+1):
	eles[i] = cu[i-1]
	
	if eles[i] == eles[i-1]:
		fodase[i] = fodase[i-1]
		
	else:
		fodase[i] = i-1
		

for i in xrange(n):
	l, r, x = map(int, raw_input().split())
	
	if eles[r] == x:
		if fodase[r] >= l:
			print fodase[r]
		else:
			print -1
			
	else:
		print r