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:
__author__ = 'Utena'
from sys import *
def input():
    return stdin.readline()
n,m=map(int,input().split())
ans1=[]
prev=[-1 for i in range(n)]
s=input().split()
for i in range(1,n):
    if s[i]!=s[i-1]:
        prev[i]=i-1
        now=i
    else:
        prev[i]=prev[i-1]
for j in range(m):
    l,r,x=input().split()
    l,r=int(l),int(r)
    if s[r-1]!=x:
        ans=r
    else:
        if prev[r-1]<l-1:ans=-1
        else:ans=prev[r-1]+1
    ans1.append(str(ans))
print('\n'.join(ans1))