Codeforces Round #463 (Div. 1 + Div. 2, combined) B. Recursive Queries （打表，前缀和）

B. Recursive Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let us define two functions f and g on positive integer numbers.

You need to process Q queries. In each query, you will be given three integers l, r and k. You need to print the number of integers xbetween l and r inclusive, such that g(x) = k.

Input

The first line of the input contains an integer Q (1 ≤ Q ≤ 2 × 105) representing the number of queries.

Q lines follow, each of which contains 3 integers l, r and k (1 ≤ l ≤ r ≤ 106, 1 ≤ k ≤ 9).

Output

For each query, print a single line containing the answer for that query.

Examples

input

Copy

4
22 73 9
45 64 6
47 55 7
2 62 4

output

1
4
0
8

input

Copy

4
82 94 6
56 67 4
28 59 9
39 74 4

output

3
1
1
5

Note

In the first example:

• g(33) = 9 as g(33) = g(3 × 3) = g(9) = 9
• g(47) = g(48) = g(60) = g(61) = 6
• There are no such integers between 47 and 55.
• g(4) = g(14) = g(22) = g(27) = g(39) = g(40) = g(41) = g(58) = 4

题意主要就是，g（x）等于各个数位和的乘积（不包含0），给出a,b,c,问ab之间有多少个g[x]=c。这个题目第一反应就是打表，做一个映射，然后再求前缀和。注意数组开大一点，不然会RE。还有一个坑点就是减的时候注意是开区间，所以后一个前缀和是a-1。

#include<bits/stdc++.h>

using namespace std;
#define maxn 2000005
int n;
int num[maxn];
int cnt[maxn][10];

int deal(int n)
{
    int sum=1;
    while(n)
    {
        if(n%10!=0)
            sum*=(n%10);
        n/=10;
    }
    return sum;
}

void create()
{
    for(int i=1; i<=maxn; i++)
    {
        int t=deal(i);
        while(t>=10)
            t=deal(t);
        num[i]=t;
        cnt[i][t]++;
    }
}


int main()
{
    create();
    for(int i=1; i<=9; i++)
        for(int j=2; j<=maxn; j++)
            cnt[j][i]+=cnt[j-1][i];
    int x,y;
    /*while(cin>>x>>y)
                cout<<cnt[x][y]<<endl;*/
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        int a,b,c;
        scanf("%d %d %d",&a,&b,&c);
        int ans=cnt[b][c]-cnt[a-1][c];    //开区间注意a-1
        printf("%d
",ans);
    }
    return 0;
}