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
_data = iter(stdin.read().split('\n'))
def input():
    while True:
        return next(_data)

n ,m = map(int, input().split())
a = list(map(int, input().split()))
ans=[]
difPre=[-1 for i in range(n)]
for i in range(1,n):
    if a[i]==a[i-1]:
        difPre[i]=difPre[i-1]
    else:
        difPre[i]=i-1

for i in range(m):
    l,r,x=map(int,input().split())
    if a[r-1]!=x:
        ans.append(str(r))
    else:
        if difPre[r-1]>=l-1:
            ans.append(str(difPre[r-1]+1))
        else:
            ans.append('-1')
print('\n'.join(ans))