cf339d Xenia and Bit Operations

Xenia and Bit Operations

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 339D

Description

Xenia the beginner programmer has a sequence a, consisting of 2ⁿ non-negative integers: a₁, a₂, ..., a_2ⁿ. 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 sequencea₁ or a₂, a₃ or a₄, ..., a_2ⁿ - 1 or a_2ⁿ, consisting of 2^n - 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 a_p = 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, ..., a_2ⁿ(0 ≤ a_i < 2³⁰). Each of the next m lines contains queries. The i-th line contains integers p_i, b_i(1 ≤ p_i ≤ 2ⁿ, 0 ≤ b_i < 2³⁰) — the i-th query.

Output

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

Sample Input

Input

2 4
1 6 3 5
1 4
3 4
1 2
1 2

Output

1
3
3
3

Hint

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

很基础的线段树

#include<iostream>
#include<stdio.h>
using namespace std;
const int maxx = <<;
int num[maxx];
int ans[maxx<<];
void build(int l,int r,int rt,int op)
{
    if(l==r)
    {
        ans[rt]=num[l];
        return;
    }
    int mid=(l+r)>>;
    build(l,mid,rt<<,-op);
    build(mid+,r,rt<<|,-op);
    if(op==) ans[rt]=ans[rt<<]|ans[rt<<|];
    else ans[rt]=ans[rt<<]^ans[rt<<|];
}
void update(int l,int r,int rt,int index,int value,int op)
{
    if(l==r&&l==index)
    {
        ans[rt]=value;
        return;
    }
    int mid=(l+r)>>;
    if(mid<index)
    {
        update(mid+,r,(rt<<|),index,value,-op);
    }
    else
    {
        update(l,mid,rt<<,index,value,-op);
    }
    if(op==) ans[rt]=ans[rt<<]|ans[rt<<|];
    else ans[rt]=ans[rt<<]^ans[rt<<|];
}
int main()
{
    //cout<<maxx+1<<endl;
    int n,m;
    scanf("%d%d",&n,&m);
    int sum=<<n;
    for(int i=;i<=sum;i++)
        scanf("%d",&num[i]);
    build(,sum,,n%);
    for(int i=;i<m;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        update(,sum,,a,b,n%);
        printf("%d\n",ans[]);
    }
    return ;
}