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 = list(map(int, input().split()))
# merge cells when a[i] == a[i+1] so we can make jumps when searching
# 1 2 1 1 3 5 - a
# 1 2 4 4 5 6 - nxt
nxt = [0] * 1000005
nxt[n] = n
for i in range(n-2,-1,-1):
    if a[i] != a[i+1]:
        nxt[i+1] = i+1
    else:
        nxt[i+1] = nxt[i+2]
'''
for i in range(7):
    print(nxt[i],end=' ')
print()
'''
for _ in range(m):
    l, r, x = map(int, sys.stdin.readline().split())
    if a[l-1] != x:
        print(l)
    elif nxt[l] < r:
        print(nxt[l]+1)
    else:
        print(-1)