Xenia and Bit Operations CodeForces

Xenia and Bit Operations CodeForces - 339D

Xenia the beginner programmer has a sequence a, consisting of 2ⁿnon-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 sequence a₁ 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 aafter 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

题意：给出一个长度为2^n序列，和几次查询，每次都将其中下标的某值改掉，之后两两或操作，得到2^(n-1)的长度后开始异或，之后在或，直至只有一个数字

题解：线段树的操作，在push_up的时候考虑一下层数是 | 还是 ^ ;

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e7+10;
const int  mod = 1e9+7;

int a[maxn];
int date[maxn];
int sum[maxn];

void push_up(int i)
{
    if(date[i]%2 == 0)
        sum[i] = sum[i<<1|1] | sum[i<<1];
    else
        sum[i] = sum[i<<1|1] ^ sum[i<<1];
}
void build(int i,int l,int r)
{
    sum[i] = 0;
    date[i << 1] = date[i<<1 | 1] = 0;
    if(l == r)
    {
        sum[i] = a[l];  date[i] = -1;   return;
    }
    int mid = (l+r) >> 1;
    build(i<<1,l,mid);
    build((i<<1)|1,mid+1,r);
    date[i] = date[i<<1]+1;
    push_up(i);
}

void update(int l,int r,int p,int d,int i)
{
    if(l == r)
    {
        sum[i] = d;
        return;
    }
    int mid = (l+r)>>1;
    if(p <= mid)
        update(l,mid,p,d,i<<1);
    else
        update(mid+1,r,p,d,i<<1|1);
    push_up(i);
}


int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int num=(1<<n);
    for(int i=1;i<=num;i++)
        scanf("%d",&a[i]);
    date[1]=1;
    build(1,1,num);
    for(int i=1;i<=m;i++)
    {
        int p,b;
        scanf("%d%d",&p,&b);
        update(1,num,p,b,1);
        printf("%d
",sum[1] );
    }
}

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