That's it for this problem. When it's gonna call this prove cedar turn Eri and turn ery takes in a list of integers. So some end number of integers, All right? And there's at least going to need to be a minimum of one of them. All right. And then we also have ah thes search algorithm, the Adam we're searching for, which is also an integer. OK, so first we wanted to find our search interval. So we'll set I equals one and j equal to end. So I is going to be the position of the first element of our search Interval and Jay will be the last element of our search internal. So at the moment, we've just initialized it to the entire list, OK? And so they will consider the case for while I is less than J minus one. Okay. And so this insurers, that builder them would stop when there's at most two elements in these search interval. We also to define a couple of new variables. Okay, so lots of definitions here, So let's let L Yeah, lower case l, uh, be equal to the floor. I plus J divided by three. Okay, we're gonna do essentially the same thing with you. This is gonna be the floor. Is there's gonna be two times I plus J divided by three. And so these are also setting positions in our list. Okay, so then we can determine in which Southern travel ex lies. Right? So if we I think about all these variables, feel lis a one and it helps you a, uh it's gonna go live to a U and then up to S O. J. Right. So were sort of crooning roughly evenly distributed intervals here. And so to find in which of these seven troubles exes in we. Well, then assign the boundaries of each search internal to the these values of the seven trolls. Right? So I say, if X is greater than or equal to hey, you then we'll set our our left interval R a. I equal to one plastic one past that, right? So they will say you plus one. It's basically an ex is larger than this element. Then we're gonna start our search at the next element cable. Then we'll do the same thing. Like what? If X is between l and you write so don't say. Well, if exes greater than or equal to a l Then in this case, we need to set. I am Jace. We need to say I is equal to l plus one. We also want our we want to stop it at Hey, you Right. So we also said j equal to you. It's that way. It's not, um, going to search from El forward, but rather just only search between l and you. No course. Lastly, it could be in this first interval, right? And so if that's the case, I would say else said j equal to L because our eyes already initialized at the 1st 1 Okay, so the thing is, this is all in this wild statement, right? So it's gonna once we narrow it down to 21 interval, it's gonna re split that interval up and do it again, and I'll split the interval. New intro found up and again and again up until we no longer have the property that eyes less than J minus one. Okay, so that will leave us with an interval containing at most two values. So continue, uh, just right side here. And so we just need to check if X is one of the values left in our search interval. Okay, so we'll say. Well, if X is equal to a I then begins at the location equal to ay right. And then we have else. If and this one we're here, we should also be in else F statement else. If X is equal to a J. The potential second item in this list then similarly will just set the location equal to J. And if neither of those cases occur, world is set the location equal to zero, which there is no zero location. So don't let us know that it does. X does not exist in in this list. And lastly, we need to return. The location is so little, Either give us a value or a zero, the location or an indication that it's not present.