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:
n, m = tuple(map(int, input().split()))
li = input().split()
a=[0]*n
for i in range(1,n):
    if li[i] != li[i-1]:
        a[i]=(i)
    else:
        a[i]=(a[i-1])
result =[0]*m
for k in range(m):
    l, r, x = input().split(); l,r=int(l),int(r)
    if li[r-1] != x: result[k]=str(r)
    elif a[r-1]>=l: result[k]=str(a[r-1])
    else: result[k]='-1'

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