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 collections
import math

n ,m = map(int, input().split())
A = list(map(int, input().split()))
ans, f = [], [0] * n
f[0] = -1
for i in range(1, n):
    if A[i] != A[i - 1]:
        f[i] = i - 1
    else:
        f[i] = f[i - 1]
for i in range(m):
    l, r, x = map(int, input().split())
    #q.append([l - 1, r - 1, x])
    #for i in range(m):
    if A[r - 1] != x:
        ans.append(r)
    elif f[r - 1] >= l - 1:
        ans.append(f[r - 1] + 1)
    else:
        ans.append(-1)
print('\n'.join(str(x) for x in ans))