Description

A number is called a Mirror number if on lateral inversion, it gives the same number i.e it looks the same in a mirror. For example 101 is a mirror number while 100 is not. 

Given two numbers a and b, find the number of mirror numbers in between them (inclusive of a and b).

Input

First line contains T, number of testcases <= 10^5. Each testcase is described in a single line containing two numbers a and b. 0 <= a<=b <= 10^44

Output

For each test case print the number of mirror numbers between a and b in a single line.

Example

Input: 3
0 10
10 20
1 4 Output: 3
1
1

演算法分析：

數位Dp求迴文串，（點選這裡），這題要求是1或0或8，且輸入為字串形式（10e44），轉化一下就行，難點在於m-1，我們需要單獨判斷m-1

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
char a[50];
int b[50];
ll dp[50][50][50];
ll dfs(int pos,int s,bool sta,int lead,int limit)
//s表示迴文串的起點,pos表示正在搜尋的位,sta表示目前構成的串是否為迴文串
{
	if(pos==-1){
		return sta;
	}
	if(!limit&&!lead&&dp[pos][s][sta]!=-1) 
	return dp[pos][s][sta];
	int up=limit?(a[pos]-'0'):9;
	ll tmp=0;
	for(int i=0;i<=up;i++)
	{
	if(i==0||i==8||i==1) 
	{
		b[pos]=i;
	if(lead==1&&i==0)     //前導0
	{
	tmp+=dfs(pos-1,s-1,sta,lead,limit&&i==up);
	}
	else 
	 { 
	    if(sta&&pos<(s+1)/2)//草稿紙看看
	 	                    //列舉後半段
	 	tmp+=dfs(pos-1,s,b[s-pos]==i,lead && i==0,limit&&i==up);
	 	else               //填充前半段
		tmp+=dfs(pos-1,s,sta,lead && i==0,limit&&i==up);
	 }
	}
	}
	if(!limit&&!lead) 
	dp[pos][s][sta]=tmp;
	return tmp; 
}
 
ll solve(string x)
{
    int pos=0;
    int len=x.length();
   for(int i=0;i<len;i++)
        a[i]=x[len-1-i];
        
    return dfs(len-1,len-1,1,true,true);//注意sta一開始要初始化為1，找了好久，因為無也是個迴文字串
}
int main()
{
	
    memset(dp,-1,sizeof(dp));
    int T;
    cin>>T;
    for(int i=1;i<=T;i++)
    {
	string n,m;
	cin>>n>>m;
	ll ans=solve(m)-solve(n);
	int len=n.length();
	int flag=1;
	for(int i=0;i<len;i++)
	{
		if((n[i]!='8'&&n[i]!='1'&&n[i]!='0')||n[i]!=n[len-i-1])
		{
			flag=0;
			break;
		}
	}
	if(flag==1) ans++;
	printf("%lld\n",ans);
    }
 
    return 0;
}