Paths in a Complete Binary Tree
Contest: Educational Codeforces Round 18
Difficulty rating: 1900
Tags: bitmasks, trees

T is a complete binary tree consisting of n vertices. It means that exactly one vertex is a root, and each vertex is either a leaf (and doesn't have children) or an inner node (and has exactly two children). All leaves of a complete binary tree have the same depth (distance from the root). So n is a number such that n + 1 is a power of 2.

In the picture you can see a complete binary tree with n = 15.

Vertices are numbered from 1 to n in a special recursive way: we recursively assign numbers to all vertices from the left subtree (if current vertex is not a leaf), then assign a number to the current vertex, and then recursively assign numbers to all vertices from the right subtree (if it exists). In the picture vertices are numbered exactly using this algorithm. It is clear that for each size of a complete binary tree exists exactly one way to give numbers to all vertices. This way of numbering is called symmetric.

You have to write a program that for given n answers q queries to the tree.

Each query consists of an integer number ui (1 ≤ ui ≤ n) and a string si, where ui is the number of vertex, and si represents the path starting from this vertex. String si doesn't contain any characters other than 'L', 'R' and 'U', which mean traverse to the left child, to the right child and to the parent, respectively. Characters from si have to be processed from left to right, considering that ui is the vertex where the path starts. If it's impossible to process a character (for example, to go to the left child of a leaf), then you have to skip it. The answer is the number of vertex where the path represented by si ends.

For example, if ui = 4 and si = «UURL», then the answer is 10.

Input format:
The first line contains two integer numbers n and q (1 ≤ n ≤ 1018, q ≥ 1). n is such that n + 1 is a power of 2.

The next 2q lines represent queries; each query consists of two consecutive lines. The first of these two lines contains ui (1 ≤ ui ≤ n), the second contains non-empty string si. si doesn't contain any characters other than 'L', 'R' and 'U'.

It is guaranteed that the sum of lengths of si (for each i such that 1 ≤ i ≤ q) doesn't exceed 105.

Output format:
Print q numbers, i-th number must be the answer to the i-th query.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Example input:
15 2
4
UURL
8
LRLLLLLLLL

Example output:
10
5

Editorial:
In this editorial x represents the number of vertex we are currently in.
Let k be the maximum integer number such that x is divisible by 2k (or the number of zeroes at the end of the binary representation of x). It is easy to prove that if k = 0, then x is a leaf; if k = 1, then both children of x are leaves, and so on. Even more, the difference between x and any of his children is exactly 2k - 1. So to traverse to the left child, we have to subtract 2k - 1 from x (if x is not a leaf), and to traverse to the right child, we add 2k - 1 to x.
How can we process traversions up? Let y be the number of the parent node. y has exactly k + 1 zeroes at the end of its binary representation, so to traverse from y to x, we need to either add or subtract 2k from y. And to traverse from x to y we also have to either subtract or add 2k to x. One of these operations will lead us to the number divisible by 2k + 1 and not divisible by 2k + 2, and we need to choose this operation.
Time complexity is $${\mathcal{O}}\left(\sum_{i=1}^{q}|s_{i}|\right)$$.