## Algorithm

no tags

Devlali numbers were an important coinage by Indian recreational mathemtician D. R. Kaprekar.

For any positive integer n, define d(n) as the sum of n and the digits of n. Eg, d(199) = 199 + 1 + 9 + 9 = 218.

For a positive number m, if there exists no positive number r such that d(r) = m, then m is a Devlali number. First few Devlali numbers are 1, 3, 5, 7, ... so on.

A prime number falling in this family is called a Devlali Prime. First few Devlali Primes are 3, 5, 7, ... so on.

### Input

First line contains integer Q

Next Q lines contain two integers A and B

### Output

print Q lines, each listing number of Devlali Primes in range [A,B] (both inclusive)

### Limits

1 <= Q <= 100000

0 <= A <= B <= 1000000

```Input
3
1 3
0 10
5 8

Output
1
3
2```

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

#define MOD                 1000000007LL
#define EPS                 1e-9
#define io                  ios_base::sync_with_stdio(false);cin.tie(NULL);
#define M_PI                3.14159265358979323846

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

const int MAXN = 1e6+5;
bool isPrime[MAXN];
bool Devlali[MAXN] = {false};

void sieve(){
isPrime[0] = isPrime[1] = 1;
for(int i = 2;i*i <= 1000000; i++){
if(isPrime[i] == 0){
if(i*1LL*i <= 1000000){
for(int j = i*i;j <= 1000000; j+=i)
isPrime[j] = 1;
}
}
}
}

int sumDig(int n){
int sum = 0;
while(n > 0){
sum += (n%10);
n /= 10;
}
return sum;
}
bool DP[MAXN];
int DP2[MAXN];
void pre(){
for(int i = 1;i <= 1000000; i++){
int sum = sumDig(i) + i;
if(sum <= 1000000){
Devlali[sum] = true;
}
}
for(int i = 1;i <= 1000000; i++){
if(Devlali[i] == false && isPrime[i] == 0)
DP[i] = true;
}
for(int i = 1;i <= 1000000; i++){
DP2[i] += DP2[i-1];
if(DP[i])
DP2[i]++;
}
}

int main(){
io;
sieve();
pre();
int q;
cin >> q;
while(q--){
int l, r;
cin >> l >> r;
if(l == 0)
cout << DP2[r] << endl;
else
cout << DP2[r] - DP2[l-1] << endl;
}
return 0;
}``````
Copy The Code &

Input

cmd
3
1 3
0 10
5 8

Output

cmd
1
3
2