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
input = sys.stdin.buffer.readline
def print(val):
    sys.stdout.write(str(val) + '\n')
def prog():
    n,m = map(int,input().split())
    a = list(map(int,input().split())) + [-1]
    intervals = [[-1,-1]]
    for i in range(n):
        if a[i] != a[i+1]:
            intervals.append([intervals[-1][1] + 1,i])
    for q in range(m):
        l,r,x = map(int,input().split())
        l -= 1
        r -= 1
        L = 0
        R = len(intervals) -1 
        while L != R:
            m = (L + R + 1)//2
            if intervals[m][0] > l:
                R = m-1
            else:
                L = m
        if a[intervals[L][0]] != x:
            print(l+1)
        elif L+1 < len(intervals) and intervals[L+1][0] <= r:
            print(intervals[L+1][0] + 1)
        else:
            print(-1)
prog()