Summer Training day6 coseforces339D 线段树、位操作

D. Xenia and Bit Operations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the beginner programmer has a sequence a, consisting of 2n non-negative integers: a1, a2, ..., a2n. Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some value v for a.

Namely, it takes several iterations to calculate value v. At the first iteration, Xenia writes a new sequence a1 or a2, a3 or a4, ..., a2n - 1 or a2n, consisting of 2n - 1 elements. In other words, she writes down the bit-wise OR of adjacent elements of sequence a. At the second iteration, Xenia writes the bitwise exclusive OR of adjacent elements of the sequence obtained after the first iteration. At the third iteration Xenia writes the bitwise OR of the adjacent elements of the sequence obtained after the second iteration. And so on; the operations of bitwise exclusive OR and bitwise OR alternate. In the end, she obtains a sequence consisting of one element, and that element is v.

Let's consider an example. Suppose that sequence a = (1, 2, 3, 4). Then let's write down all the transformations (1, 2, 3, 4) → (1 or 2 = 3, 3 or 4 = 7) → (3 xor 7 = 4). The result is v = 4.

You are given Xenia's initial sequence. But to calculate value v for a given sequence would be too easy, so you are given additional mqueries. Each query is a pair of integers p, b. Query p, b means that you need to perform the assignment ap = b. After each query, you need to print the new value v for the new sequence a.

Input

The first line contains two integers n and m (1 ≤ n ≤ 17, 1 ≤ m ≤ 105). The next line contains 2n integers a1, a2, ..., a2n (0 ≤ ai < 230). Each of the next m lines contains queries. The i-th line contains integers pi, bi (1 ≤ pi ≤ 2n, 0 ≤ bi < 230) — the i-th query.

Output

Print m integers — the i-th integer denotes value v for sequence a after the i-th query.

Examples

input

output

Note

For more information on the bit operations, you can follow this link: http://en.wikipedia.org/wiki/Bitwise_operation

题解：把题目给出的所有元素构建一颗线段树，对于更新操作，类似于线段树的操作，从下往上处理线段树的子区间，用当前层的深度来确定对每一层实施什么样的擦操作。然后pushup操作更新当前区间。时间复杂度O(logn)

代码：

#include <iostream>
#include <cstdio>
using namespace std;
int a[1<<18];
int n,m;
int dfs(int x,int dep,int l,int r){if(l == r){return a[x];}int mid = (l + r) / 2;int ls = dfs(2*x,dep+1,l,mid);int rs = dfs(2*x + 1,dep + 1,mid + 1,r);if((n - dep) % 2 == 0){//ora[x] = ls | rs;}else{a[x] = ls ^ rs;}return a[x];
}
int main(){scanf("%d%d",&n,&m);for(int i = 0;i < (1<<n);i++){int tmp;scanf("%d",&tmp);a[(1<<n)+i] = tmp;}dfs(1,1,1,(1<<n));while(m--){int p,b;scanf("%d%d",&p,&b);int fa = (1<<n) + p - 1;a[fa] = b;int dep = 0;while(fa != 1){fa = fa / 2;if(dep % 2 == 0)a[fa] = a[fa*2] | a[fa*2+1];elsea[fa] = a[fa*2] ^ a[fa*2+1];dep ++;}printf("%d\n",a[1]);}return 0;
}

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