The problem can be cracked down as "Find the number of non-negative integer solutions for the equation $$$a_1 + a_2 + \ldots + a_n = k$$$, with $$$0 \le a_1, a_2, \ldots, a_n \le m$$$."

Without regarding the complexity, there is an inclusion-exclusion-based solution as of following:

• Denote $$$S_{mask}$$$ ($$$0 \le mask < 2^n$$$) as the number of integer solutions for the above equation, with $$$mask$$$ being the bitmask saying which of the unknown variable $$$a_i$$$ would have the criteria $$$a_i \ge m+1$$$ considered. For example, if $$$n = 3$$$, $$$S_5$$$ will count the number of solutions with only $$$a_1$$$ and $$$a_3$$$ having their lower-bound limited.
• The answer would be $$$\displaystyle \sum_{mask = 0}^{2^n - 1} (-1)^{bitcount(mask)} \times S_{mask}$$$, with $$$bitcount(mask)$$$ being the number of $$$1$$$ bit in the binary notation of $$$mask$$$.

One can obviously see that this solution is way too slow. However, the trick is that if a variable is bounded, no matter which one, the lower-bound limit would always be $$$m+1$$$. This makes all variables being equal, and without loss of generality, we can merge all the $$$mask$$$ value with the same $$$bitcount(mask)$$$ to calculate at one instance.

From here, our solution will become:

• Iterate value $$$i$$$ — the number of variables having their lower-bound limited in the equation — from $$$0$$$ to $$$n$$$.
• One can see that such equation can be transformed into such with no variables being bounded anyhow (except of the mandatory of being non-negative) by subtracting from both sides all the lowerbound values, thus, as a result, if $$$k - i(m+1) < 0$$$, obviously no solution would exist.
• Otherwise, denote the number of solutions as $$$C_i$$$, we can see that $$$C_i = \binom{n+k-i(m+1)-1}{n-1} \times \binom{n}{i}$$$, with $$$\binom{n+k-i(m+1)-1}{n-1}$$$ being the number of solutions for each equation of our considering form, $$$\binom{n}{i}$$$ being the number of equations possible (in other words, the number of ways to pick $$$i$$$ variables to have their lower-bound raised to $$$m+1$$$).
• Add $$$(-1)^i \times C_i$$$ to the final answer.

Last but not least, the module in this problem ($$$10^6+7$$$) is unusual — it is a compound of two primes, $$$29$$$ and $$$34483$$$. Therefore, we should calculate separately with each prime as a independent modulo (please take note that from $$$1$$$ to $$$10^5$$$ there might include some multiples of those primes as well), and merge the answer using either Chinese Remainder Theorem or brute force (since the final module is small, one can try out all possible values from $$$0$$$ to $$$10^6+6$$$).

Total complexity would be $$$\mathcal{O}(n \log(mod))$$$, or $$$\mathcal{O}(mod)$$$ if you decide to brute force the remainder after calculating the individual steps. ;)