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 PYTHON3):

Solution:
from sys import stdin

n, m = [int(i) for i in stdin.readline().split()]
ori = [int(i) for i in stdin.readline().split()]
pre = [-1] * n
ans = list()
for i in range(n-1):
	if ori[i] != ori[i+1]:
		pre[i+1] = i
	else:
		pre[i+1] = pre[i]
for i in range(m):
	l, r, x = [int(i) for i in stdin.readline().split()]
	tem = r - 1
	if ori[tem] != x: 
		ans.append(str(r))
	else:
		if pre[tem] < l-1:
			ans.append('-1')
		else:
			ans.append(str(pre[tem]+1))

print('\n'.join(ans))