D. Xenia and Bit Operations Codeforces Round 197 (Div. 2)

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博客围绕C++编程，讲述Xenia对由2n个非负整数组成的序列a进行二进制操作计算值v，操作包括相邻元素的位或和位异或且交替进行。还给出初始序列及m个查询，每次查询更新序列元素后需输出新的v值，解题用线段树存图思想。

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.

Examples

input

Copy

output

Copy

Note

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

题目大致意思：进行n次二进制操作，先相邻的数或（|）操作，然后相邻的数异或（^）操作。

在这里用线段树存图的思想，1是一整段的值，2是前一半操作完的值，3是后一半操作完的值……

void gx(ll x){
	ll y = 1;
	while(x){
		if(y&1)v[x] = v[x*2] | v[x*2+1];
		else v[x] = v[x*2] ^ v[x*2+1];
		x >>= 1;
		y++;
	}
	return;
}

可以发现存图所需要的空间正好是1<<n,也就是 $2^{n}$ 空间。

所以我们正常存图就只能在1<<n到1<<(n+1),也就是 $2^{n}$ 到 $2^{n+1}$ 这里。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"

const ll N = 1e6+7;
ll n,m;
ll v[N];

void gx(ll x){
	ll y = 1;
	while(x){
		if(y&1)v[x] = v[x*2] | v[x*2+1];
		else v[x] = v[x*2] ^ v[x*2+1];
		x >>= 1;
		y++;
	}
	return;
}

void solve(){
	cin >> n >> m;
	for(ll i = 1<<n ; i < 1<<(n+1) ; i ++){
		cin >> v[i];
		gx(i/2);
	}
	while(m --){
		ll x,y;
		cin >> x >> y;
		v[(1<<n)+x-1] = y;
		gx((1<<n)+x-1 >> 1);
		cout << v[1] << endl;
	}
	return;
}

int main(){
	ll t=1;//cin >> t;
	while(t--)solve();
	return 0;
}