#include <iostream>
using namespace std;
 
int main() {
    int n;
    while (cin >> n) {
        // 找到不大于n的最大2的幂次
        int power = 1;
        while (power * 2 <= n) {
            power *= 2;
        }
 
        // 最后剩下的数字是 2*(n-power) + 1
        // 或者更简单：n的二进制表示去掉最高位后的值乘以2
        int result = 2 * (n - power) + 1;
        cout << result << endl;
    }
    return 0;
}

I2luY2x1ZGUgPGlvc3RyZWFtPgp1c2luZyBuYW1lc3BhY2Ugc3RkOwoKaW50IG1haW4oKSB7CiAgICBpbnQgbjsKICAgIHdoaWxlIChjaW4gPj4gbikgewogICAgICAgIC8vIOaJvuWIsOS4jeWkp+S6jm7nmoTmnIDlpKcy55qE5bmC5qyhCiAgICAgICAgaW50IHBvd2VyID0gMTsKICAgICAgICB3aGlsZSAocG93ZXIgKiAyIDw9IG4pIHsKICAgICAgICAgICAgcG93ZXIgKj0gMjsKICAgICAgICB9CiAgICAgICAgCiAgICAgICAgLy8g5pyA5ZCO5Ymp5LiL55qE5pWw5a2X5pivIDIqKG4tcG93ZXIpICsgMQogICAgICAgIC8vIOaIluiAheabtOeugOWNle+8mm7nmoTkuozov5vliLbooajnpLrljrvmjonmnIDpq5jkvY3lkI7nmoTlgLzkuZjku6UyCiAgICAgICAgaW50IHJlc3VsdCA9IDIgKiAobiAtIHBvd2VyKSArIDE7CiAgICAgICAgY291dCA8PCByZXN1bHQgPDwgZW5kbDsKICAgIH0KICAgIHJldHVybiAwOwp9

NTAw

500