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, input().split())
a = [0] + list(map(int, input().split()))
b = [n+1]*(n+1)
for i in range(n-1, 0, -1):
    b[i] = i+1 if a[i] != a[i+1] else b[i+1]

ans = [-1]*m
for i in range(m):
    l, r, x = map(int, sys.stdin.readline().split())
    if a[l] != x:
        ans[i] = l
    elif b[l] <= r:
        ans[i] = b[l]

print(*ans, sep='\n')