The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point. 
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them? 

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description. 

The input terminates by end of file marker. 

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

3
1
50
500

Sample Output

0
1
15

Hint

From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.

题意：和上一篇类似，稍微一改就是，这是要49，，给你一个区间，问里面包含49的数有多少个

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll dp[20][2];
int a[20];
ll dfs(int pos,int pre,int sta,int limit)
{
    if(pos==-1)return 1;
    if(!limit&&dp[pos][sta]!=-1) return dp[pos][sta];
    int up=limit?a[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
    {
        if(pre==4&&i==9)continue;
        ans+=dfs(pos-1,i,i==4,limit&&i==a[pos]);
    }
    if(!limit) dp[pos][sta]=ans;
    return ans;
}
ll solve(ll x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,1,0,1);
}
int main()
{
    int T;
    ll x;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,-1,sizeof(dp));
        scanf("%lld",&x);
        printf("%lld\n",x-solve(x)+(ll)1);
    }
    return 0;
}