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:
# -*- coding: utf-8 -*-
"""
Created on Tue May 17 15:26:55 2016

@author: Alex
"""
from sys import stdin

n,m = [int(i) for i in input().split()]
l = [int(i) for i in stdin.readline().split()]
prev = [-1 for i in range(n)]
re = []
for i in range(n-1):
    if l[i] != l[i+1]:
        prev[i+1] = i
    else:
        prev[i+1] = prev[i]
for _ in range(m):
    le,ri,xi = [int(i) for i in stdin.readline().split()]
    i = ri-1
    if l[i]!=xi:
        re.append(str(ri))
    else:
        if prev[i]<le-1:
            re.append('-1')
        else:
            re.append(str(prev[i]+1))
print('\n'.join(re))