Farmer John is sending a cow to compete in the International Bovine Olympiad, but he's too lazy to write problems for a contest, so he does the next most reasonable thing: a fighting tournament! Farmer John has (2^n) cows standing in a line, with the (i)-th cow having a skill level of (a_i). Each cow starts in a stack containing only itself, and the skill level of the stack is equal to the XOR-sum(^{\text{∗}}) of the skill levels of all the cows in it. For example, if the stack consists of cows (1, 3, 9) from bottom to top, then the skill level of the stack is (1 \oplus 3 \oplus 9 = 11). The following process repeats until there is only one stack. Every stack in an odd position ((1)st, (3)rd, etc.) starts a fight with the stack to its right. For example, the (1)st stack will fight with the (2)nd stack, the (3)rd stack will fight with the (4)th stack, etc. The stack with the higher skill level will win the match, with the left-most stack winning in case of a tie. The winning stack will jump on top of the losing stack and shift itself over such that there are no gaps left by the defeated stacks. In order to make this more exciting, Farmer John created (q) potions, such that the (i)-th potion sets the skill level of the cow who drinks it to (c_i). Farmer John wants to test his potions on the cows, and so he gives the (i)-th potion to cow (b_i) and then runs the tournament. For each trial, Farmer John wants to know how many cows are above the cow that was given the potion in the final stack. Farmer John's potions wear off quickly, so when the tournament ends, the cow that was given the potion will have its skill level return to what it originally was . In other words, all queries are independent. (^{\text{∗}})The XOR-sum of an array (x_1, x_2, \ldots, x_y) is equal to (x_1 \oplus x_2 \oplus x_3 \ldots x_{y-1} \oplus x_y) where (\oplus) denotes the bitwise XOR operation . The fi