You've got an array a, consisting of n integers a1, a2, ..., an. You are allowed to perform two operations on this array:

1. Calculate the sum of current array elements on the segment [l, r], that is, count value al + al + 1 + ... + ar.
2. Apply the xor operation with a given number x
    to each array element on the segment [l, r], that is, execute . This operation changes exactly r - l + 1 array elements.

Expression  means applying bitwise xor operation to numbers x and y. The given operation exists in all modern programming languages, for example in language C++ and Java it is marked as "^", in Pascal — as "xor".

You've got a list of m operations of the indicated type. Your task is to perform all given operations, for each sum query you should print the result you get.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the size of the array. The second line contains space-separated integers a

The third line contains integer m (1 ≤ m ≤ 5·104) — the number of operations with the array. The i-th of the following m lines first contains an integer ti (1 ≤ ti ≤ 2) — the type of the i-th query. If ti = 1, then this is the query of the sum, if ti = 2, then this is the query to change array elements. If the i-th operation is of type 1, then next follow two integers li, ri (1 ≤ li ≤ ri ≤ n). If the i-th operation is of type 2, then next follow three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106). The numbers on the lines are separated by single spaces.

Output

For each query of type 1 print in a single line the sum of numbers on the given segment. Print the answers to the queries in the order in which the queries go in the input.

兩個操作

1：求[ L,R ] 的和；

2：[ L,R ] 的每一個數a[ i ] ^ x;

關鍵是加法運算對異或運算不滿足分配律，就是a^c+b^c != (a+b)^c;

怎麼辦了？那就把a二進位制化，維護區間每個二進位制位的數目；

當x的某一位為1，區間中所有a[ i ]所有為1的要變為0，為0的變為1，然而當其為0時，a[i]對應位是0或者1對答案均無影響 ；

所以維護區間中每個二進位制位上1的數目即可，最後按權加和答案就行了；

異或和會爆int；

#include<iostream>
#include<vector>
#include<set>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
const int MOD=1e9+7;
const int maxn=1e5+7;
const double ESP=1e-8;
int n;
int a[maxn];
struct node
{
    int dig[25];///每一位1的數目
    int lazy;
}b[maxn<<2];
void up(int rt)
{
    for(int i=0;i<22;++i)
        b[rt].dig[i]=b[rt<<1].dig[i]+b[rt<<1|1].dig[i];
}
void down(int l,int r,int rt)
{
    if(b[rt].lazy)
    {
        b[rt<<1].lazy^=b[rt].lazy;
        b[rt<<1|1].lazy^=b[rt].lazy;
        int mid=(l+r)>>1;
        for(int i=0;i<22;++i)
        {
            if(b[rt].lazy&1<<i)
            {
                b[rt<<1].dig[i]=mid-l+1-b[rt<<1].dig[i];
                b[rt<<1|1].dig[i]=r-mid-b[rt<<1|1].dig[i];
            }
        }
        b[rt].lazy=0;

    }
}
void build(int l,int r,int rt)
{
    b[rt].lazy=0;
    if(l==r)
    {
        for(int i=0;i<22;++i)
            if(a[l]&1<<i) b[rt].dig[i]=1;
            else b[rt].dig[i]=0;
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    up(rt);
}
void change(int l,int r,int rt,int ll,int rr,int x)
{
    if(ll<=l&&r<=rr)
    {
        b[rt].lazy^=x;
        for(int i=0;i<22;++i)
            if(x&1<<i) b[rt].dig[i]=r-l+1-b[rt].dig[i];
        return ;
    }
    down(l,r,rt);
    int mid=(l+r)>>1;
    if(ll<=mid) change(l,mid,rt<<1,ll,rr,x);
    if(mid<rr)  change(mid+1,r,rt<<1|1,ll,rr,x);
    up(rt);
}
long long query(int l,int r,int rt,int ll,int rr)
{
    long long res=0;
    if(ll<=l&&r<=rr)
    {
        for(int i=0;i<22;++i)
            res+=(long long)b[rt].dig[i]*(1<<i);
        return res;
    }
    down(l,r,rt);
    int mid=(l+r)>>1;
    if(ll<=mid) res+=query(l,mid,rt<<1,ll,rr);
    if(mid<rr)  res+=query(mid+1,r,rt<<1|1,ll,rr);
    up(rt);
    return res;

}
int q;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%d",&a[i]);
    scanf("%d",&q);
    int op,l,r,x;
    build(1,n,1);
    while(q--)
    {
        scanf("%d",&op);
        if(op==1)
        {
            scanf("%d%d",&l,&r);
            cout<<query(1,n,1,l,r)<<endl;
        }
        else
        {
            scanf("%d%d%d",&l,&r,&x);
            change(1,n,1,l,r,x);
        }
    }
}