segment tree cf197D Xenia and Bit Operations

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.

Examples

input

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

output

1
3
3
3

题意： 给你2^n个数，让你每两个进行一次按位或操作，得到一个新的序列，（新的序列长度变为2^(n-1)). 接下来每两个进行按位异或，然后再按位或，反复循环至到数列只剩下一个数，输出那个数。

分析：线段树，只需要单点更新即可。要判断异或和或的次序。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
#include<iomanip>
using namespace std;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
typedef long long ll;
const int maxn = 1 << 18; 
int n,m;
int segtree[maxn],num[maxn];
void build(int l,int r,int pos,bool key){   //建树
    if(r - l == 1){
        segtree[pos] = num[l];
        return;
    } 
    int mid = (l+r)/2;
    build(l,mid,pos*2+1,!key);
    build(mid,r,pos*2+2,!key);
    if(key == true){
        segtree[pos] = segtree[pos*2+1] | segtree[pos*2+2];
    }
    else segtree[pos] = segtree[pos*2+1] ^ segtree[pos*2+2];
}
int change(int x,int y){
    x += (1<<n)-1;	//找到叶结点
    segtree[x] = y;
    int k = 1;
    while(x > 0){           //向上更新
        x = (x-1) / 2;
        if(k&1) segtree[x] = segtree[x*2+1] | segtree[x*2+2];
        else segtree[x] = segtree[x*2+1] ^ segtree[x*2+2];
        k++;
    }
}
int main()
{
    while(cin >> n >> m){
        for(int i=0;i<(1<<n);i++){
            cin >> num[i];
        }
        bool k = false;
        if(n&1) k = true;  //判断 异或还是或 的操作
        build(0,(1<<n),0,k);
        for(int i=0;i<m;i++){
            int x,y;
            cin >> x >> y;
            x--;
            change(x,y);
            cout << segtree[0] << endl;
        }
    }
}