 somehow these Israeli Jews are not as tough as I was told there would be. Alright, next up, if you ever thought that Brunstein is kind of tight, well there's a Brunstein bound, it's almost tight, it is tight. So, Miro Nandi is next. Okay, thank you. So, I hope you don't know Dan, but some of you do not know his bound. So, first, let me explain to you what his bound is. Okay, so it is the upper bound for our Hirukhwariya advantage, your Persian advantage for something oil known as the oilman card. So, how do you look like? You have a polyhash, you have a message. So, you just apply every position of the message as a coefficient of the polynomial and key is your variable. And then we have some norms which should not repeat and we apply some random parameters on the block side for it and then we add. So, that's the whiteboard factor of the indicator. And what Brunstein proved in the year 2005 that the maximum poetry advantage for any such indicator can be at most, let's call it B and Q which has something like some expression like this. So, let me first understand what it means. So, we have a L over root point factor and other than that, we have 1 minus Q over root point or something. So, it may look something weird. Some people sometimes call this the Parthi bound or maybe something other. So, from this expression probably we cannot interpret one bound root line. So, let's have a different root of the bound. Actually, this same bound can be written as a different form. So, this L over root point is there times Q square over root point plus 1. So, this Q square over root point plus 1 is a popular form of the bound. It will be called the Parthi bound. But it is not Parthi bound. So, for example, if we look at S1, the K, which is just a Q is over and the root point factor. So, Q is the number of map queries. For the time being, you can ignore them. So, it is the number of passes blocks you can take it 1 in this talk. So, if you take Q is root point by 2, that's the Parthi complexity. And the bound says that you cannot get the advantage more than 1.65 over root point. So, that's the maximum advantage you can get. That's the bound. And if you just make a random guess of the key, then you get advantage 1 over root point. So, maximum you can get 1.65 times advantage over random guess. So, what you can say that the Parthi bound is already known quick type, all adversaries making Parthi bound complexities. And the case two, if you want to go beyond Parthi, if you want to make queries more than 2 point n by 2, for example, even if you make queries less than 2 point n by 2, still the Parthi bound, the Parthi bound is very close to zero. So, you have to go something like 2 point n by 2 point n by 2 to achieve something non-negative to that one. So, that's the interpretation of the Parthi bound. So, if you have attended the yesterday talk by Byrd Pendel, there is a paper in the URK here. So, Loose and Pendel, they provided a data. And the complexity of that is 2 point n by 2, ok. So, the Buster already proved that you cannot achieve more than 1 point 6 over 2 point n. So, it's already known to be tight. And if you calculate the advantage, you will say you will get at 1 point 4 over 2 point n. So, it's something like to me it's like a worse than guessing a 1 bit of a key. So, if you have a n bit key, if you just guess 1 bit key, then the advantage is 2 over 2 point n. But you are not even guessing 1 bit, 1 bit, it makes sense. Ok, so, this is one thing and that's why Optimite is already known. So, it's nothing new among the all Parthi bound complexity adversaries. And that does not say anything when the adversary is making your Parthi complexity. Ok, so, this is the summary of this in their paper. So, I have something new is this. So, if you make queries this will be 2 point n by 2. So, if you make queries square root of n times 2 point n by 2, then actually you can have advantage which is like something like half. So, you can actually actually advantage like a significant advantage like half. If you make queries square root of n times 2 point n by 2, and that actually shows that Parthi bound is tight. Because Parthi bound shows that if you make queries less than 2 point n by 2, then the advantage is almost zero. And if you make 2 point n by 2, then the advantage can be significant and that's what I have shown. And to show this, we have two types of advantage. So, one case you can choose the message randomly. And in that case, the analysis is fairly simple. And you can prove that the advantage is half. But sometimes the message may not be chosen randomly, or you can assume any distribution of the message. Ok, so take any non-random or any distribution of the message, as long as they are distinct, you can actually prove the advantage is still half. Ok, but that analysis will do complex. So, if you want to find some details on this, then what we have to do, we have to come to the end of our work. Thank you.