One integer number x is called "Mountain Number" if:

(1) x>0 and x is an integer;

(2) Assume x=a[0]a[1]...a[len-2]a[len-1](0≤a[i]≤9, a[0] is positive). Any a[2i+1] is larger or equal to a[2i] and a[2i+2](if exists).

For example, 111, 132, 893, 7 are "Mountain Number" while 123, 10, 76889 are not "Mountain Number".

Now you are given L and R, how many "Mountain Number" can be found between L and R (inclusive) ?

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only two integers L and R (1≤L≤R≤1,000,000,000).

Output

For each test case, output the number of "Mountain Number" between L and R in a single line.

Sample Input

3
1 10
1 100
1 1000

Sample Output

9
54
384

題意：

如果一個>0的整數x，滿足a[2*i+1] >= a[2*i]和a[2*i+2]，則這個數為Mountain Number。

給出L， R，求區間[L, R]有多少個Mountain Number。

思路：

數位DP，判斷當前是偶數位還是奇數位(從0開始)，

如果是偶數位，那麼它要比前一個數的值小，

如果是奇數位，那麼它要比前一個數的值大。

不明白為什麼開long  long 就答案錯誤

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
int a[30];
int dp[30][15][15];
int dfs(int pos,int pre,int sta,bool limit){
 
	if(pos<0){
		return 1;
	}
	if(!limit&&dp[pos][pre][sta]!=-1) 
	return dp[pos][pre][sta];
	int up=limit?a[pos]:9;
	int tmp=0;
	for(int i=0;i<=up;i++)
	{
	 if(sta && i<=pre) tmp+=dfs(pos-1,i,0,limit && i==up);
	 if(!sta && i>=pre) tmp+=dfs(pos-1,i,1,limit && i==up);
	}
	if(!limit) 
	dp[pos][pre][sta]=tmp;
	return tmp; 
}

int solve(int x)
{
	memset(a,0,sizeof(a));
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,9,1,1);
}
int main()
{
	
    memset(dp,-1,sizeof(dp));
    int n,m;
    int T;
    cin>>T;
    while(T--)
    {        
    	scanf("%d%d",&n,&m);
        printf("%d\n",solve(m)-solve(n-1));
    }
    return 0;
}