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 input().split()]
a = [int(i) for i in stdin.readline().split()]
re = []
nxt = [n for i in range(n)]
for i in range(n-2,-1,-1):
    if a[i] != a[i+1]:
        nxt[i] = i+1
    else:
        nxt[i] = nxt[i+1]

for _ in range(m):
    l,r,x = [int(i) for i in stdin.readline().split()]
    i = l - 1
    if a[i] != x:
        re.append(str(i+1))
    else:
        if nxt[i] >= r:
            re.append('-1')
        else:
            re.append(str(nxt[i]+1))

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