CodeForces 893B Beautiful Divisors

Description：

Recently Luba learned about a special kind of numbers that she calls beautiful numbers. The number is called beautiful iff its binary representation consists ofk + 1 consecutive ones, and then k consecutive zeroes.

Some examples of beautiful numbers:

• 1₂ (1₁₀);
• 110₂ (6₁₀);
• 1111000₂ (120₁₀);
• 111110000₂ (496₁₀).

More formally, the number is beautiful iff there exists some positive integer k such that the number is equal to (2^k - 1) * (2^k - 1).

Luba has got an integer number n, and she wants to find its greatest beautiful divisor. Help her to find it!

Input

The only line of input contains one number n (1 ≤ n ≤ 10⁵) — the number Luba has got.

Output

Output one number — the greatest beautiful divisor of Luba's number. It is obvious that the answer always exists.

Example

Input

3

Output

1

Input

992

Output

496

题目大意：

给定一个n，求小于等于n的最大的漂亮数字。漂亮数字的定义是二进制下前k个数字都为1， 后k + 1个数字都为0的十进制数字。

解题思路：

因为要求最大的， 我们只要从小于等于n的漂亮数字中从后往前枚举看是否能被整除即可。

代码：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

/*
 *ios::sync_with_stdio(false);
 */

typedef long long ll;
typedef unsigned long long ull;
const int dir[5][2] = {0, 1, 0, -1, 1, 0, -1, 0, 0, 0};
const ll ll_inf = 0x7fffffff;
const int inf = 0x3f3f3f;
const int mod = (int)1e9 + 7;
const int Max = (int) 1e5 + 7;

int ans, cur;

int main() {
    int n;
    ans = 1, cur = 1;
    scanf("%d", &n);
    while (ans <= n) {
        cur++;
        int x = 0;
        for (int i = cur - 1; i <= cur + cur - 2; ++i) {
            x += (1 << i);
        }
        ans = x;
    }
    for (int i = cur; i >= 1; --i) {
        int temp = 0;
        for (int j = i - 1; j <= i + i - 2; ++j) {
            temp += (1 << j);
        }
        if (n >= temp && n % temp == 0) {
            printf("%d\n", temp);
            return 0;
        }
    }
}