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:
import sys


n, m = map(int, sys.stdin.readline().split())
a = list(map(int, sys.stdin.readline().split()))
assert(len(a) == n)

prev = [-1] * n
l = r = 0
while l < n:
    while r < n and a[l] == a[r]:
        r += 1
    # [l, r) --> a[l]
    for i in range(l, r):
        prev[i] = l - 1
    l = r

for i in range(m):
    l, r, x = map(int, sys.stdin.readline().split())
    l -= 1; r -= 1
    if a[r] != x:
        print(r + 1)
    elif prev[r] == -1 or prev[r] < l:
        print(-1)
    else:
        print(prev[r] + 1)