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:
#!/usr/bin/env python3

def solver(ifs):
    n, m = tuple(map(int, ifs.readline().split()))
    a = tuple(map(int, ifs.readline().split()))
    b = [0,] * len(a)
    c = [0,] * len(a)
    for i in range(1, n):
        if a[i] == a[i-1]:
            b[i] = b[i-1] + 1
    for i in range(n-2, -1, -1):
        if a[i] == a[i+1]:
            c[i] = c[i+1] + 1
    for i in range(m):
        L, R, x = tuple(map(int, ifs.readline().split()))
        L, R = L-1, R-1
        left = L + c[L] + 1
        right = R - b[R] - 1
        res = -1
        if a[L] != x:
            res = L + 1
        elif a[R] != x:
            res = R + 1
        elif left <= R:
            res = left + 1
        elif L <= right:
            res = right + 1
        print(res)

def main():
    import sys
    if sys.version_info.major == 3:
        from io import StringIO as StreamIO
    else:
         from io import BytesIO as StreamIO
    
    with StreamIO(sys.stdin.read()) as ifs, StreamIO() as ofs:
        _stdout = sys.stdout
        sys.stdout = ofs
        solver(ifs)
        sys.stdout = _stdout
        sys.stdout.write(ofs.getvalue())
    return 0

if __name__ == '__main__':
    main()