Bob watches TV every day. He always sets the volume of his TV to?bb. However, today he is angry to find out someone has changed the volume to?aa. Of course, Bob has a remote control that can change the volume.

There are six buttons (?5,?2,?1,+1,+2,+5?5,?2,?1,+1,+2,+5) on the control, which in one press can either increase or decrease the current volume by?11,?22, or?55. The volume can be arbitrarily large, but can never be negative. In other words, Bob cannot press the button if it causes the volume to be lower than?00.

As Bob is so angry, he wants to change the volume to?bb?using as few button presses as possible. However, he forgets how to do such simple calculations, so he asks you for help. Write a program that given?aa?and?bb, finds the minimum number of presses to change the TV volume from?aa?to?bb.

Input

Each test contains multiple test cases. The first line contains the number of test cases?TT?(1≤T≤10001≤T≤1000). Then the descriptions of the test cases follow.

Each test case consists of one line containing two integers?aa?and?bb?(0≤a,b≤1090≤a,b≤109)?— the current volume and Bob's desired volume, respectively.

Output

For each test case, output a single integer?— the minimum number of presses to change the TV volume from?aa?to?bb. If Bob does not need to change the volume (i.e.?a=ba=b), then print?00.

Example

Input

3 4 0 5 14 3 9

Output

2 3 2

Note

In the first example, Bob can press the??2?2?button twice to reach?00. Note that Bob can not press??5?5?when the volume is?44?since it will make the volume negative.

In the second example, one of the optimal ways for Bob is to press the?+5+5?twice, then press??1?1?once.

In the last example, Bob can press the?+5+5?once, then press?+1+1.

能選5的時候選5，不能選5的時候選2，不能選2的時候選1。

驗證可行性，如果有個大于等于5的數(shù)，不選5那么至少要選兩個2一個1，甚至更多，所以直接貪心，完事。

#include<bits/stdc++.h> #define MAXM 100000000 #define MAXN 1000005 using namespace std; int main() {int a,b,t;cin>>t;while(t--){cin>>a>>b;int ans=abs(a-b);int c=ans%5;int sum=(ans-c)/5;ans=c%2;sum+=(c-ans)/2+ans;cout<<sum<<endl;} }

?

總結(jié)

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