Absorption in naive floating-point summation

In IEEE-754 double precision, you sum, left to right, one value of $10^{16}$ followed by exactly $10^{6}$ (one million) values of $1.0$. The exact mathematical total is $10^{16} + 10^{6} = 10{,}000{,}000{,}001{,}000{,}000$. What total does the naive running-sum loop actually return? (A double has a 52-bit fraction, so around $10^{16}$ its representable values are spaced $2$ apart.)