#include<stdio.h>

int main(void){
int a,b,c,i,point;
point=0;
scanf("%d %d %d",&a,&b,&c);
for(i=a;i<=b;i++){
if(c%i==0)point++;
}

printf("%d\n",point);


return 0;

}

### Prompt

Develop a solution in Cpp to the problem described below:

How Many Divisors?

Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.

Constraints

* 1 ≤ a, b, c ≤ 10000
* a ≤ b

Input

Three integers a, b and c are given in a line separated by a single space.

Output

Print the number of divisors in a line.

Example

Input

5 14 80


Output

3

### Response

```cpp
#include<stdio.h>

int main(void){
int a,b,c,i,point;
point=0;
scanf("%d %d %d",&a,&b,&c);
for(i=a;i<=b;i++){
if(c%i==0)point++;
}

printf("%d\n",point);


return 0;

}

```