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 stdin.readline().split()]
a = [int(i) for i in stdin.readline().split()]
b = [None] * n
b[0] = -1
ans = [None] * m
for i in range(1,len(a)):
    if a[i] == a[i - 1]:
        b[i] = b[i - 1]
    else:
        b[i] = i - 1
for i in range(m):
    l,r,x = [int(j) for j in stdin.readline().split()]
    k = r - 1
    if a[k] != x:
        ans[i] = str(k + 1)
    elif b[k] + 1 >= l:
        ans[i] = str(b[k] + 1)
    else:
        ans[i] = ('-1')

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