F. Foo Fighterstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given n objects. Each object has two integer properties: val_i — its price — and mask_i. It is guaranteed that the sum of all prices is initially non-zero.You want to select a positive integer s. All objects will be modified after that. The i-th object will be modified using the following procedure:   Consider mask_i and s in binary notation,  Compute the bitwise AND of s and mask_i (s \,\&\, mask_i),  If (s \,\&\, mask_i) contains an odd number of ones, replace the val_i with -val_i. Otherwise do nothing with the i-th object. You need to find such an integer s that when the modification above is done the sum of all prices changes sign (if it was negative, it should become positive, and vice-versa; it is not allowed for it to become zero). The absolute value of the sum can be arbitrary.InputThe first line contains a single integer n (1 \leq n \leq 3 \cdot 10^5) — the number of objects.The i-th of next n lines contains integers val_i and mask_i (-10^9 \leq val_i \leq 10^9, 1 \le mask_i \le 2^{62} - 1) — the price of the object and its mask.It is guaranteed that the sum of val_i is initially non-zero.OutputPrint an integer s (1 \le s \le 2^{62} - 1), such that if we modify the objects as described above, the sign of the sum of val_i changes its sign.If there are multiple such s, print any of them. One can show that there is always at least one valid s.ExamplesInput
5
17 206
-6 117
-2 151
9 93
6 117
Output
64
Input
1
1 1
Output
1
NoteIn the first test sample all objects will change their prices except for the object with mask 151. So their total sum will change its sign: initially 24, after modifications — -28.In the second test sample the only object will change its price. So the total sum will change its sign. 
5
17 206
-6 117
-2 151
9 93
6 117

64 
['bitmasks', 'constructive algorithms', '*2700']