# Product of first N factorials in C++

Given a number N, the task is to find the product of first N factorials modulo by 1000000007. . Factorial implies when we find the product of all the numbers below that number including that number and is denoted by ! (Exclamation sign), For example − 4! = 4x3x2x1 = 24.

So, we have to find a product of n factorials and modulo by 1000000007..

Constraint

1 ≤ N ≤ 1e6.

Input

n = 9

Output

27

Explanation

1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! Mod (1e9 + 7) = 27

Input

n = 3

Output

12

Explanation

1! * 2! * 3! mod (1e9 +7) = 12

## Approach used below is as follows to solve the problem

• Find the factorial recursively from i = 1 till n and product all the factorials

• Mod the product of all the factorials by 1e9 +7

• Return the result.

## Algorithm

In Fucntion long long int mulmod(long long int x, long long int y, long long int mod)
Step 1→ Declare and Initialize result as 0
Step 2→ Set x as x % mod
Step 3→ While y > 0
If y % 2 == 1 then,
Set result as (result + x) % mod
Set x as (x * 2) % mod
Set y as y/ 2
Step 4→ return (result % mod)
In Function long long int nfactprod(long long int num)
Step 1→ Declare and Initialize product with 1 and fact with 1
Step 2→ Declare and Initialize MOD as (1e9 + 7)
Step 3→ For i = 1 and i <= num and i++
Set fact as (call function mulmod(fact, i, MOD))
Set product as (call function mulmod(product, fact, MOD))
If product == 0 then,
Return 0
Step 4→ Return product
In Function int main()
Step 1→ Declare and Initialize num = 3
Step 2→ Print the result by calling (nfactprod(num))
Stop

## Example

Live Demo

#include <stdio.h>
long long int mulmod(long long int x, long long int y, long long int mod){
long long int result = 0;
x = x % mod;
while (y > 0) {
// add x where y is odd.
if (y % 2 == 1)
result = (result + x) % mod;
// Multiply x with 2
x = (x * 2) % mod;
// Divide y by 2
y /= 2;
}
return result % mod;
}
long long int nfactprod(long long int num){
// Initialize product and fact with 1
long long int product = 1, fact = 1;
long long int MOD = 1e9 + 7;
for (int i = 1; i <= num; i++) {
// to find factorial for every iteration
fact = mulmod(fact, i, MOD);
// product of first i factorials
product = mulmod(product, fact, MOD);
//when product divisible by MOD return 0
if (product == 0)
return 0;
}
return product;
}
int main(){
long long int num = 3;
printf("%lld \n", (nfactprod(num)));
return 0;
}

## Output

If run the above code it will generate the following output −

12