For simplicity, let the array be

~[a, b, c, d]~

After one second:

~[a, a+b, a+b+c, a+b+c+d]~

Two seconds:

~[a, 2a+b, 3a+2b+c, 4a+3b+2c+d]~

Three seconds:

~[a, 3a+b, 6a+3b+c, 10a+6b+3c+d]~

It's easy to recognize these coefficients as diagonals in Pascal's Triangle, which are related to Combinations.

To understand why this is the case, consider the grid where each row ~t~ is the array ~a~ at time ~t~:

~{a_1}_1, {a_1}_2, {a_1}_3, {a_1}_4~

~{a_2}_1, {a_2}_2, {a_2}_3, {a_2}_4~

~{a_3}_1, {a_3}_2, {a_3}_3, {a_3}_4~

~{a_4}_1, {a_4}_2, {a_4}_3, {a_4}_4~

How much will ~{a_1}_1~ contribute to the sum in ~{a_4}_4~? Each path of only right and down moves between ~{a_2}_1~ and ~{a_4}_4~ represents one "propagation" of ~{a_1}_1~ to ~{a_4}_4~ from repeated prefix-summing. It's a bit abstract, but moving right is choosing to continue the propagation along the row in a specific frame of time, while moving down is choosing to move one second forward.

So, each term ~a_i~'s contribution to the final sum is ~{(N+T-i) \choose T}~.

Now, how do we calculate this mod ~1e9+7~?

Recall that ~n \choose x~ ~=~ ~\frac {n!} {x!(n-x)}~

The multiplication is quite straightforward, and the factorial can be precomputed. However, to perform the modular division, say, to divide by ~x~, you have to multiply by the modular inverse of ~x~ in mod ~1e9+7~.

One way to find this modular inverse is using Fermat's Little Theorem, which states that ~a^{p-1} = 1~ ~mod~ ~p~ for primes ~p~. By extension, ~a^{p-2} = {\frac {1} {a}}~ ~mod~ ~p~. That means to divide by ~x~ in mod ~1e9+7~, it is equivalent to multiplying by ~x^{p-2}~ ~mod~ ~p~. To calculate this power, the following binary exponentiation method can be used:

long long fastPow(long long base, long long expo, long long mdl){
    long long ret = 1;
    while (expo > 0){
        if (expo&1) ret = (ret*base) % mdl;
        base = (base*base) % mdl;
        expo >>= 1;
    }
    return ret;
}