Choose the smallest correct upper bound on the solution to the following recurrence: t(1) = 1t(1)=1 and t(n) \le t([ \sqrt{n} ]) + 1t(n)≤t([ n ])+1 for n> 1. here [x] denotes the "floor" function, which rounds down to the nearest integer. (note that the master method does not apply.)