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, ..., 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 mqueries. 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.

Sample test(s)

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

题目大意：给定一个数组，然后交叉进行下面的操作

• In other words, she writes down the bit-wise OR of adjacent elements of sequence a.（相邻元素同或）
• Xenia writes the bitwise exclusive OR of adjacent elements of the sequence obtained after the first iteration（相邻元素异或）

               当然了，还有一个操作：

               Query p, b means that you need to perform the assignment ap = b. 

               对于Query p，b我们知道这题可以用线段树解决。因为修改了ap ，那么后面的值也要相应的做出改变...

下面是代码：

#include 
#define maxn 200200

int sum[maxn<<2];
int A[maxn];

//如果op为1,则为 OR
//如果op为0,则为 ~OR
void build(int l,int r,int rt,int op)
{
    if(l == r)
    {
        sum[rt] = A[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(l,mid,rt<<1,1-op);
    build(mid+1,r,rt<<1|1,1-op);
    if(op) sum[rt] = sum[rt<<1] | sum[rt<<1|1];
    else sum[rt] = sum[rt<<1] ^ sum[rt<<1|1];
}

void update(int p,int l,int r,int rt,int v,int op)
{
    if(l == r)
    {
        sum[rt] = v;
        return ;
    }
    int mid = (l + r) >> 1;
    if(p > mid) update(p,mid+1,r,rt<<1|1,v,1-op);
    else if(p <= mid) update(p,l,mid,rt<<1,v,1-op);
    if(op) sum[rt] = sum[rt<<1] | sum[rt<<1|1];
    else sum[rt] = sum[rt<<1] ^ sum[rt<<1|1];
}
int main()
{
    int n,m,cnt;
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= (1<