Editorial

Transform the expression:

x + y + x \cdot y = y \cdot (x + 1) + (x + 1) - 1 = (x + 1) \cdot (y + 1) - 1

After removing the numbers $x$ and $y$ from the list, the number $(x + 1) \cdot (y + 1) - 1$ will be inserted.

If you further remove $(x + 1) \cdot (y + 1) - 1$ and $z$ from the list, the next number will be inserted:

((x + 1) \cdot (y + 1) - 1 + 1) \cdot (z + 1) - 1 = (x + 1) \cdot (y + 1) \cdot (z + 1) - 1

By analogy, it can be stated that if initially

L = a_{1}, a_{2}, ..., a_{n},

then at the end there will be a number $(a_{1} + 1) \cdot (a_{2} + 1) \cdot ... \cdot (a_{n} + 1) - 1$ .

Since initially $L$ contains integers from $1$ to $n$ , at the end there will be a number

(1 + 1) \cdot (2 + 1) \cdot ... \cdot (n + 1) -1 = (n + 1)! - 1

Algorithm realization

Declare the constants.

#define MOD 1000000007
#define MAX 1000002

The input contains several test cases. Declare an array $p$ such that $p [i] = i!$ .

long long p[MAX];

Read the number of test cases $t es t s$ . Compute the factorials of numbers, and store $p [i] = i!$ .

scanf("%d", &tests);
p[1] = 1;
for (i = 2; i < MAX; i++)
  p[i] = (p[i - 1] * i) % MOD;

Process the test cases. For each input value of $n$ print $(n + 1)! - 1$ .

while (tests--)
{
  scanf("%d", &n);
  printf("%lld\n", p[n + 1] - 1);
}

Python realization

Declare the constants.

MOD = 1000000007
MAX = 1000002

Compute the factorials of numbers, and store $p [i] = i!$ .

p = [1] * MAX
for i in range(2, MAX):
  p[i] = (p[i - 1] * i) % MOD

Read the number of test cases $t es t s$ .

tests = int(input())

Process the test cases. For each input value of $n$ print $(n + 1)! - 1$ .

for _ in range(tests):
  n = int(input())
  print(p[n + 1] - 1)