Codeforces Round #197 (Div. 2) -- D. Xenia and Bit Operations（线段树）

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, ..., 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 m queries. 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.

Examples

Input

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

Output

1
3
3
3

Note

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

大体题意：

给你n和m

给你2^n个数和m 询问！

计算出最终的结果v

其中v 第一次是相邻的数进or 运算，第二次 进行xor运算，，，，依次类推！

思路：

很明显是线段树题目，与一般的线段树有些不同，这个只要求出最终的值即可！可以把线段树的第一个节点记为最终答案，深度是奇数时就进行xor运算，深度是偶数时进行or运算

在一点不同就是这个线段树有两个数字就该计算了，父结点是id的话，那么递归构造左儿子id<<1  和右儿子(id<<1) ^ 1 。

最后输出T[1]即可！

#include
#include
#include
using namespace std;
const int maxn = 200000 + 10;
int a[maxn];
struct Segment_tree{
    struct Node{
        int s,t,val,dep;
        void get(int dd,int l,int r){
            s = l,t = r,dep = dd,val = a[l];
        }
    }T[maxn<<2];
    void build(int id,int dep,int l,int r){
        T[id].get(dep,l,r);
        if (l == r)return ;
        int mid = (l+r) >> 1;
        build(id<<1,dep-1,l,mid);
        build((id<<1)^1,dep-1,mid+1,r);
        pushnode(T[id],T[id<<1],T[(id<<1)^1]);
    }
    void pushnode(Node & fa,Node &lc,Node & rc){
        if (fa.dep % 2 == 1)fa.val = lc.val | rc.val;
        else fa.val = lc.val^rc.val;
    }
    void update(int id,int pos,int val){
        if (T[id].s == T[id].t){
            T[id].val = val;
            return ;
        }
        int mid = (T[id].s + T[id].t) >> 1;
        if (pos <= mid)update(id<<1,pos,val);
        else update((id<<1)^1,pos,val);
        pushnode(T[id],T[id<<1],T[(id<<1)^1]);
    }
}st;
int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    int dep = n;
    n = (1<

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