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 m queries. 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.

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

題意：
輸入n，m
以及2^n個數字組成的陣列
接下來是m行
每行輸入p，b，將a[p]的值改成b，然後計算數值，計算規則如下：

第一次計算b1 = a[1] | a[2] , b2 = a[ 3 ] | a[ 4 ] , b3 = a[ 5 ] | a[ 6 ] , b4 = a[ 7 ] | a[ 8 ]
第二次計算c1 = b1^b2 , c2 = b3^b4
第四次計算v = c1 | c2
輸出v

簡單來說 就是 | ^ 這兩種運算子的交替運算，然後輸出最終的數值

#include <bits/stdc++.h>
using namespace std;
const int MAXN = (1<<17)+10;
int a[MAXN];
int ff[MAXN<<2];
int sum[MAXN<<2];
void pb(int rt)
{
    if(ff[rt]%2==0)
    {
        sum[rt]=sum[rt<<1|1]|sum[rt<<1];
    }
    else sum[rt]=sum[rt<<1|1]^sum[rt<<1];
}

void build(int l,int r,int rt)
{
    sum[rt]=0;
    ff[rt<<1]=ff[rt<<1|1]=0;
    if(l==r)
    {
        sum[rt]=a[l];
        ff[rt]=-1;
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    ff[rt]=ff[rt<<1]+1;
    pb(rt);
}

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

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