题意：给定两个音量，需要从l调到r，每次只能向上一下或者向下（1，2，4，2^n）下（连续向下的情况下），每次可以停歇或向上来打断向下的连续性，最少需要多少步？

思路：当l小于r时显然答案是r-l，当l>r时那么不停地下降到r附近，此时只需要在r的上方和下方挑选最少的步数即可。注意向上也可以打断向下的连续性。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=105;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin)
#define OUT freopen("out.txt","w",stdout)int read() {char ch;while (ch = getchar(), !isdigit(ch));int res = ch - '0';while (ch = getchar(), isdigit(ch))res = res * 10 + ch - '0';return res;
}
ll powm(ll a,ll n){ll ans=1;while (n){if (n&1) ans=ans*a%mod;a=a*a%mod;n>>=1;}return ans%mod;
}
//++++++++++++密++++++++++++++封++++++++++++++++++++线ll n,m;
ll sol(ll n,ll stp){if (n<=m) return max(0LL,m-n-stp);ll nn=n,p=1,sp=0;while (n-m>0){n-=p;p<<=1;sp++;if (n==m) return sp;}sp+=min(sol(n+p/2,++stp),sol(max(0LL,n),stp));return sp;
}
int main()
{int T=read();ft(ca,1,T){scanf("%lld%lld",&n,&m);cout<<sol(n,0)<<endl;}
}