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 *

n, m = [int(t) for t in input().split()]

vec = []
pre = []
for idx, val in enumerate([int(t) for t in stdin.readline().split()]):
    vec.append(val)

    last = -1 if idx == 0 else (idx-1 if vec[idx-1] != val else pre[idx-1])
    pre.append(last)

ans = []
for q in range(m):
    l, r, x = [int(t) for t in stdin.readline().split()]

    l -= 1
    r -= 1

    if vec[l] == x and vec[r] == x:
        ans.append(str(-1 if pre[r] < l else pre[r]+1))
    else:
        ans.append(str(l+1 if vec[l] != x else r+1))

print("\n".join(ans))
# PYPY !!