Game on Sum (Hard Version)
Contest: Codeforces Round 767 (Div. 1)
Difficulty rating: 2400
Tags: combinatorics, dp, games, math

This is the hard version of the problem. The difference is the constraints on $$$n$$$, $$$m$$$ and $$$t$$$. You can make hacks only if all versions of the problem are solved.

Alice and Bob are given the numbers $$$n$$$, $$$m$$$ and $$$k$$$, and play a game as follows:

The game has a score that Alice tries to maximize, and Bob tries to minimize. The score is initially $$$0$$$. The game consists of $$$n$$$ turns. Each turn, Alice picks a real number from $$$0$$$ to $$$k$$$ (inclusive) which Bob either adds to or subtracts from the score of the game. But throughout the game, Bob has to choose to add at least $$$m$$$ out of the $$$n$$$ turns.

Bob gets to know which number Alice picked before deciding whether to add or subtract the number from the score, and Alice gets to know whether Bob added or subtracted the number for the previous turn before picking the number for the current turn (except on the first turn since there was no previous turn).

If Alice and Bob play optimally, what will the final score of the game be?

Input format:
The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. The description of test cases follows.

Each test case consists of a single line containing the three integers, $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1 \le m \le n \le 10^6, 0 \le k < 10^9 + 7$$$) — the number of turns, how many of those turns Bob has to add, and the biggest number Alice can choose, respectively.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output format:
For each test case output a single integer number — the score of the optimal game modulo $$$10^9 + 7$$$.

Formally, let $$$M = 10^9 + 7$$$. It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{p}{q}$$$, where $$$p$$$ and $$$q$$$ are integers and $$$q \not \equiv 0 \pmod{M}$$$. Output the integer equal to $$$p \cdot q^{-1} \bmod M$$$. In other words, output such an integer $$$x$$$ that $$$0 \le x < M$$$ and $$$x \cdot q \equiv p \pmod{M}$$$.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Note:
In the first test case, the entire game has $$$3$$$ turns, and since $$$m = 3$$$, Bob has to add in each of them. Therefore Alice should pick the biggest number she can, which is $$$k = 2$$$, every turn.

In the third test case, Alice has a strategy to guarantee a score of $$$\frac{75}{8} \equiv 375000012 \pmod{10^9 + 7}$$$.

In the fourth test case, Alice has a strategy to guarantee a score of $$$\frac{45}{2} \equiv 500000026 \pmod{10^9 + 7}$$$.

Example input:
7
3 3 2
2 1 10
6 3 10
6 4 10
100 1 1
4 4 0
69 4 20

Example output:
6
5
375000012
500000026
958557139
0
49735962

Editorial:
We have base cases
• $$$DP[i][0] = 0$$$
• $$$DP[i][i] = k\cdot i$$$
And transition $$$DP[i][j] = (DP[i-1][j-1]+DP[i-1][j])/2$$$
Check the explanation for the easy version to see why.
This $$$DP$$$ can be optimized by looking at contributions from the base cases.
If we draw the $$$DP$$$ states on a grid and ignore the division by $$$2$$$ in the transition, we can see that the number of times states $$$DP[i][i]$$$ contributes to state $$$DP[n][m]$$$ is the number of paths from $$$(i, j)$$$ to $$$(n, m)$$$ in the grid such that at each step, both $$$i$$$ and $$$j$$$ increase, or only $$$j$$$ increases, except we have to exclude paths that go through other base cases. The number of such paths is $$$n-i-1 \choose m-j$$$. Since the number of steps in all of these paths is the same, we can account for the division by $$$2$$$ in each transition by dividing by $$$2^{n-i}$$$ in the end.
To find the value of $$$DP[n][m]$$$ we sum the contribution form every base case $$$DP[i][i]$$$ for $$$1 \leq i \leq n$$$.