Description

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

output

input

output

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：

给定一个表达式，根据表达式求值。先输入q，代表有q次询问，接下来q行，分别为l,r,k。表示从l到r有多少个值为k的数字，输出一个整数

思路：

看数据量，q和l,r都很大，所以如果直接暴力会超时。所以要先求出每个数的值是多少。这里通过递归或者循环都能求出来，本人不习惯递归，使用循环求出。然后用二维数组，直接求出从1-i有多少个符合条件的数，二位数组的i代表从1到i有多少符合条件的，j代表的是结果，也就是k。所以求l到r之间等于k的就是ans[r][k] - ans[l-1][k]。具体看代码。

代码：

#include <bits/stdc++.h>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <cmath>

#include <iostream>

#include <algorithm>

#include <string>

#include <queue>

#include <stack>

#include <map>

#include <set>

#define IO ios::sync_with_stdio(false);\

    cin.tie();\

    cout.tie();

typedef long long LL;

const long long INF = 0x3f3f3f3f;

const long long mod = 1e9+;

const double PI = acos(-1.0);

const int maxn = ;

const char week[][]= {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};

const char month[][]= {"Janurary","February","March","April","May","June","July",

                           "August","September","October","November","December"

                          };

const int daym[][] = {{, , , , , , , , , , , , },

    {, , , , , , , , , , , , }

};

const int dir4[][] = {{, }, {, }, {-, }, {, -}};

const int dir8[][] = {{, }, {, }, {-, }, {, -}, {, }, {-, -}, {, -}, {-, }};

using namespace std;

using namespace std;

int n, k, ans[maxn][], q, l, r;

int a[maxn];

void init()//将1-maxn的值求出来

{

    for(int i=; i<=maxn; i++)

    {

        int sum=;

        int ii=i;

        while(ii)

        {

            if(ii%!=)

                sum=sum*(ii%);

            ii/=;

        }

        int sum1=;

        while()

        {

            if(sum<)

            {

                a[i]=sum;

                break;

            }

            while(sum)

            {

                if(sum%!=)

                    sum1=sum1*(sum%);

                sum/=;

            }

            sum=sum1;

            sum1=;

        }

    }

}

int main()

{

    init();

    for (int i = ; i <= maxn; i++)//将每个数的结果加上

    {

        ans[i][a[i]]++;

    }

    for (int i = ; i <= ; i++)

        for (int j = ; j <= maxn; j++)//将从1到j的等于i的数求出来。

            ans[j][i] += ans[j-][i];

    cin >> q;

    while (q--)

    {

        cin >> l>> r>> k;

        cout <<ans[r][k] - ans[l-][k]<<endl;

    }

}

巴特西

前缀和：CodeForces 932B Recursive Queries

Description

Input

Output

Examples

Note

题意：

思路：

代码：

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