D. K-good

Problem - D - Codeforces

D. K-good

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We say that a positive integer nn is kk-good for some positive integer kk if nn can be expressed as a sum of kk positive integers which give kk distinct remainders when divided by kk.

Given a positive integer nn, find some k≥2k≥2 so that nn is kk-good or tell that such a kk does not exist.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1051≤t≤105) — the number of test cases.

Each test case consists of one line with an integer nn (2≤n≤10182≤n≤1018).

Output

For each test case, print a line with a value of kk such that nn is kk-good (k≥2k≥2), or −1−1 if nn is not kk-good for any kk. If there are multiple valid values of kk, you can print any of them.

Example

input

Copy

5
2
4
6
15
20

output

Copy

-1
-1
3
3
5

Note

66 is a 33-good number since it can be expressed as a sum of 33 numbers which give different remainders when divided by 33: 6=1+2+36=1+2+3.

1515 is also a 33-good number since 15=1+5+915=1+5+9 and 1,5,91,5,9 give different remainders when divided by 33.

2020 is a 55-good number since 20=2+3+4+5+620=2+3+4+5+6 and 2,3,4,5,62,3,4,5,6 give different remainders when divided by 55.

 CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes!) Editorial - Codeforces

 相当于k1不满足条件的时候，k2是可行解

这里最初假定k是偶数，则k+1为奇数，这样k和（k+1）/2都不是n的因子，因为（k+1）/2不是整数

所以说当且仅当

2n=k （mod k）是显然成立的

说的不是很清楚 

import sys
import math
#import numpy as np
from collections import defaultdict
#sys.stdin = open("in.txt", "r")
 
T = int(input())
for _ in range(T):
    n = int(input())
    k=1
    while n%2==0:
        n=n//2
        k*=2
    if n==1:
        print('-1')
    else:
        print(min(n,2*k))