Xenia and Bit Operations CodeForces

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.

Example

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

题解：很明显的线段树，单点更新，只是要给每个结点标记一下，总算线段树入门啦！

#include<bits/stdc++.h>
#define lson l , m , rt << 1
#define rson m+1,r , rt << 1 | 1
using namespace std;

const int maxn=2e5;

int n,m,p,b;
int cal[20],x[4*maxn],mark[4*maxn];

int get(int k){
    int tem=1;
    for(int i=1;i<=k;i++) tem*=2;
    return tem;
}

void Inite(){
    for(int i=1;i<=20;i++) cal[i]=get(i);
}

void Bookup(int rt){
    int sonl=rt<<1;
    int sonr=rt<<1|1;
    if(mark[sonl] && mark[sonr]) mark[rt]=0;
    if(!mark[sonl]&&!mark[sonr]) mark[rt]=1;
}

void Pushup(int rt){
    int sonl=rt<<1;
    int sonr=rt<<1|1;
    //cout<<"ROOT: "<<rt<<endl;
    //cout<<mark[rt]<<" "<<x[sonl]<<" "<<x[sonr]<<endl;
    if(mark[rt]==1) x[rt]=x[sonl]|x[sonr];
    if(mark[rt]==0) x[rt]=x[sonl]^x[sonr];
}

void Build(int l,int r,int rt){
    if(l==r){
        scanf("%d",&x[rt]);
        return;
    }
    int m=(l+r)>>1;
    Build(lson);
    Build(rson);
    Bookup(rt);
    Pushup(rt);
}

void Update(int l,int r,int rt){
    if(l==r){
        x[rt]=b;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m) Update(lson);
    else Update(rson);
    Pushup(rt);
}

int main()
{   Inite();
    scanf("%d%d",&n,&m);
    Build(1,cal[n],1);
    while(m--){
        scanf("%d%d",&p,&b);
        Update(1,cal[n],1);
        printf("%d
",x[1]);
    }
    return 0;
}