Alice is playing a game with permutations of size N.

She selects a random permutation P of size N and a random index i (1≤ i ≤ N);
She keeps incrementing the index i until:
- The next element is greater than the current element (P_{(i+1)} > P_{i}), or;
- She reaches the last index (i = N).

Find the [expected] number of indices that Alice will visit throughout the game.

It can be shown that this expected value can be expressed as $a fraction \frac{P}{Q}$, where P and Q are coprime integers, P ≥ 0, Q > 0 and Q is coprime with 10^{9} + 7. You should compute $P \cdot Q^{-1}$ \% (10^{9} + 7), where Q^{-1} denotes the multiplicative inverse of Q modulo 10^{9}+7.

Note that a permutation of size N consists of all integers from 1 to N, where each element is present exactly once.

------ Input Format ------

• The first line of input will contain a single integer T, denoting the number of test cases.
• Each test case consists of single lines of input containing an integer N, the size of permutation.

------ Output Format ------

For each test case, print in a single line, the expected number of indices that Alice will visit throughout the game.

It can be shown that this expected value can be expressed as $a fraction \frac{P}{Q}$, where P and Q are coprime integers, P ≥ 0, Q > 0 and Q is coprime with 10^{9} + 7. You should compute $P \cdot Q^{-1}$ \% (10^{9} + 7), where Q^{-1} denotes the multiplicative inverse of Q modulo 10^{9}+7.

------ Constraints ------

$1 ≤ T ≤ 10^{3}$
$1 ≤ N ≤ 2\cdot 10^{5}$
- The sum of $N$ over all test cases won't exceed $2\cdot 10^{5}$.

----- Sample Input 1 ------
3
1
2
3

----- Sample Output 1 ------
1
250000003
388888893

----- explanation 1 ------
Test case $1$: There is only one possible way to select a permutation of size $1$ and an index from the permutation $1$. Thus, we choose $P = \{1\}$ and $i = 1$. The number of indices covered after this choice is $1$.

Test case $2$: The possible cases are:
- $P = \{1, 2\}$ and $i = 1$: We cannot increment the index $i=1$ as $P_{2} > P_{1}$. Thus, number of indices covered is $1$.
- $P = \{1, 2\}$ and $i = 2$: We cannot increment the index $i=2$ as it is the last index. Thus, number of indices covered is $1$.
- $P = \{2, 1\}$ and $i = 1$: We can increment the index $i=1$ as $P_{2} < P_{1}$. We reach index $i = 2$. Now we cannot increment the index. Thus, number of indices covered is $2$.
- $P = \{2, 1\}$ and $i = 2$: We cannot increment the index $i=2$ as it is the last index. Thus, number of indices covered is $1$.

Thus, the expected value of number of indices visited is $\frac{3}{4}\cdot 1 + \frac{1}{4}\cdot 2 = \frac{5}{4}$.

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