#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using indexed_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define int long long
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define eb emplace_back
#define ins insert
#define F first
#define S second
#define MAXN 1000005
#define LOG (int)log2(MAXN)+1
const int INF = 1e18;
const int MOD = 998244353;
const double EPS = 1e-8;
/*
a xor (b xor c) = a xor b xor c
a xor b xor a = b
||
\/
a1 xor a2 xor (a1 xor a2) = a1 xor a2 xor a1 xor a2 = 0
xor - ^
a1 ^ a2 ^ a3 ^ (a1^a2) ^ (a2 ^ a3) ^ (a1^a2^a3) = a1 ^ a2 ^ a3 ^ a1 ^ a2 ^ a2 ^ a3 ^ a1 ^ a2 ^ a3 =
= a1 ^ a3
query(l, r) = ?
essentially it's xor of all sliding windows in [l,r].
let's see how many times the i-th element occurs in a sliding window of length k
freq[i-th element] = min(i-l+1, r-i+1, k)
We need to know the sum for all k, then the i-th element participates in xor if the sum is odd for it,
and does not participate if it is even.
Observation 1:
cnt[i-th element] = (r-i+1)*(i-l+1)
So we can answer queries in O(n*q) 12+18+25 = 55p.
*/
void solve(){
int n, q;
cin >> n >> q;
vector<int> a(n+1);
for(int i = 1; i <= n; i++) cin >> a[i];
int t, l, r;
while(q--){
cin >> t >> l >> r;
if(t==1){
a[l] = r;
}else{
int res = 0, cnt;
for(int i = l; i <= r; i++){
cnt = (r-i+1)*(i-l+1);
if(cnt&1) res ^= a[i];
}
cout << res << '\n';
}
}
}
signed main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int t = 1;
//cin >> t;
for(int i = 1; i <= t; i++){
//cout << "Case " << i << ": ";
solve();
}
return 0;
}
