Two best friends Serozha and Gena play a game.
Initially there is one pile consisting of _n_ stones on the table. During one move one pile should be taken and divided into an arbitrary number of piles consisting of _a_1 > _a_2 > ... > _a__k_ > 0 stones. The piles should meet the condition _a_1 - _a_2 = _a_2 - _a_3 = ... = _a__k_ - 1 - _a__k_ = 1. Naturally, the number of piles _k_ should be no less than two.
The friends play in turns. The player who cannot make a move loses. Serozha makes the first move. Who will win if both players play in the optimal way?
## Input
The single line contains a single integer _n_ (1 ≤ _n_ ≤ 105).
## Output
If Serozha wins, print _k_, which represents the minimal number of piles into which he can split the initial one during the first move in order to win the game.
If Gena wins, print "-1" (without the quotes).
[samples]
两个最好的朋友 Serozha 和 Gena 玩一个游戏。
最初桌上有且仅有一堆包含 #cf_span[n] 个石子的石堆。每一步，玩家必须选择一堆石子，并将其拆分为任意数量的石堆，这些石堆的石子数满足 #cf_span[a1 > a2 > ... > ak > 0]，且必须满足条件 #cf_span[a1 - a2 = a2 - a3 = ... = ak - 1 - ak = 1]。显然，石堆的数量 #cf_span[k] 至少为 2。
两位玩家轮流进行操作，无法进行操作的玩家输掉游戏。Serozha 先手。如果双方都采取最优策略，谁会获胜？
输入仅一行，包含一个整数 #cf_span[n] (#cf_span[1 ≤ n ≤ 105])。
如果 Serozha 获胜，请输出 #cf_span[k]，表示他在第一步中将初始石堆拆分成的最少石堆数量以确保获胜。
如果 Gena 获胜，请输出 "-1"（不含引号）。
## Input
输入仅一行，包含一个整数 #cf_span[n] (#cf_span[1 ≤ n ≤ 105])。
## Output
如果 Serozha 获胜，请输出 #cf_span[k]，表示他在第一步中将初始石堆拆分成的最少石堆数量以确保获胜。如果 Gena 获胜，请输出 "-1"（不含引号）。
[samples]
Let $ n $ be the initial number of stones.
A move consists of replacing one pile of size $ n $ with $ k \geq 2 $ piles of sizes $ a_1 > a_2 > \dots > a_k > 0 $ such that:
$$
a_i - a_{i+1} = 1 \quad \text{for all } 1 \leq i < k.
$$
This implies the pile sizes form a consecutive decreasing sequence:
$$
a_i = a_1 - (i - 1), \quad \text{so} \quad a_k = a_1 - (k - 1).
$$
The total number of stones is:
$$
\sum_{i=1}^k a_i = \sum_{j=0}^{k-1} (a_1 - j) = k a_1 - \frac{(k-1)k}{2}.
$$
Set equal to $ n $:
$$
k a_1 - \frac{k(k-1)}{2} = n \quad \Rightarrow \quad a_1 = \frac{n + \frac{k(k-1)}{2}}{k} = \frac{2n + k(k-1)}{2k}.
$$
For $ a_1 $ to be a positive integer, $ \frac{2n + k(k-1)}{2k} \in \mathbb{Z}^+ $, and since $ a_k = a_1 - (k - 1) > 0 $, we require:
$$
a_1 \geq k \quad \Rightarrow \quad \frac{2n + k(k-1)}{2k} \geq k \quad \Rightarrow \quad 2n + k(k-1) \geq 2k^2 \quad \Rightarrow \quad 2n \geq k^2 + k.
$$
Thus, a valid move exists for $ k \geq 2 $ if and only if:
$$
k(k+1) \leq 2n \quad \text{and} \quad \frac{2n + k(k-1)}{2k} \in \mathbb{Z}.
$$
Let $ G(n) $ denote the Grundy number (nimber) of a pile of size $ n $.
Base case: $ G(1) = 0 $ (no move possible).
For $ n \geq 2 $:
$$
G(n) = \mathrm{mex} \left\{ \bigoplus_{i=1}^k G(a_i) \,\middle|\, \text{valid split into } k \geq 2 \text{ piles with consecutive decreasing sizes summing to } n \right\}.
$$
But since the split produces piles of sizes $ a_1, a_1 - 1, \dots, a_1 - k + 1 $, and the game is impartial with independent piles, the Grundy value of the resulting position is:
$$
\bigoplus_{j=0}^{k-1} G(a_1 - j).
$$
We are to compute $ G(n) $ for $ 1 \leq n \leq 10^5 $.
If $ G(n) = 0 $, then the first player (Serozha) loses → output "-1".
If $ G(n) \ne 0 $, then Serozha wins. We must find the **minimal** $ k \geq 2 $ such that:
1. $ k(k+1) \leq 2n $,
2. $ \frac{2n + k(k-1)}{2k} \in \mathbb{Z} $,
3. The resulting position has Grundy number 0 (i.e., the move leads to a losing position for Gena).
That is, find minimal $ k \geq 2 $ satisfying (1), (2), and:
$$
\bigoplus_{j=0}^{k-1} G\left( \frac{2n + k(k-1)}{2k} - j \right) = 0.
$$
---
**Formal Output Requirement:**
Given $ n \in [1, 10^5] $:
- If $ G(n) = 0 $, output $-1$.
- Else, output the minimal $ k \geq 2 $ such that:
$$
\begin{cases}
k(k+1) \leq 2n, \\
\dfrac{2n + k(k-1)}{2k} \in \mathbb{Z}^+, \\
\displaystyle \bigoplus_{j=0}^{k-1} G\left( \left\lfloor \frac{2n + k(k-1)}{2k} \right\rfloor - j \right) = 0.
\end{cases}
$$
(Note: The floor is redundant since condition 2 ensures integrality.)