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_2^n - 1 or a_2ⁿ, 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 a_p = b. After each query, you need to print the new value v for the new sequence a.

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.

Print m integers — the i-th integer denotes value v for sequence a after the i-th query.

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

1
3
3
3

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {if(Head == Tail) {int l = fread(buffer, 1, BufferSize, stdin);Tail = (Head = buffer) + l;}return *Head++;
}
int read() {int x = 0, f = 1; char c = getchar();while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }return x * f;
}#define maxn 20
#define maxm 1048576
int n, q, ToT, A[maxm], fa[maxm];int main() {n = read(); q = read();for(int i = 0; i < (1 << n); i++) A[i] = read();memset(fa, -1, sizeof(fa));ToT = 1 << n; bool cur = 0;int nl = 0, nr = (1 << n) - 1;for(int i = 1; i <= n; i++, cur ^= 1) {int nnl = -1, nnr = -1;for(int j = nl; j < nr; j += 2) {if(cur) A[ToT++] = A[j] ^ A[j^1], fa[j] = fa[j^1] = ToT - 1;else A[ToT++] = A[j] | A[j^1], fa[j] = fa[j^1] = ToT - 1;if(nnl < 0) nnl = ToT;nnr = ToT;}nl = nnl; nr = nnr;}
//  printf("%d\n", ToT);while(q--) {int p = read() - 1; A[p] = read();cur = 0;while(fa[p] >= 0) {if(cur) A[fa[p]] = A[p] ^ A[p^1];else A[fa[p]] = A[p] | A[p^1];p = fa[p]; cur ^= 1;}printf("%d\n", A[ToT-1]);}return 0;
}