Problem:
Score : 200 points

Problem Statement

Joisino is about to compete in the final round of a certain programming competition.
In this contest, there are N problems, numbered 1 through N.
Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N).

Also, there are M kinds of drinks offered to the contestants, numbered 1 through M.
If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds.
It does not affect the time to solve the other problems.

A contestant is allowed to take exactly one of the drinks before the start of the contest.
For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink.
Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems.
Your task is to write a program to calculate it instead of her.

Constraints

All input values are integers.

1≦N≦100

1≦T_i≦10^5

1≦M≦100

1≦P_i≦N

1≦X_i≦10^5

Input

The input is given from Standard Input in the following format:

N
T_1 T_2 ... T_N
M
P_1 X_1
P_2 X_2
:
P_M X_M

Output

For each drink, calculate how many seconds it takes Joisino to solve all the problems if she takes that drink, and print the results, one per line.

Sample Input 1

3
2 1 4
2
1 1
2 3

Sample Output 1

6
9

If Joisino takes drink 1, the time it takes her to solve each problem will be 1, 1 and 4 seconds, respectively, totaling 6 seconds.

If Joisino takes drink 2, the time it takes her to solve each problem will be 2, 3 and 4 seconds, respectively, totaling 9 seconds.

Sample Input 2

5
7 2 3 8 5
3
4 2
1 7
4 13

Sample Output 2

19
25
30

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
ID: your_id_here
PROG: test
LANG: C++11
*/
#include <bits/stdc++.h>
using namespace std;
typedef  long long ll;
double pi=3.1415926535898;
//double pi=acos(-1.0);
#define scarr(a,s,e) for(int i=s;i<=int(e);i++) scanf("%d ",&a[i]);
#define prarr(a,s,e) for(int i=s;i<=int(e);i++) printf("%d ",a[i]);
#define speed      ios::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
#define rng(i,a,b) for(int i=a;i>=int(b);i--)
#define rep(i,a,b) for(int i=a;i<=int(b);++i)
#define lop(i,n)   for(int i=0;i<int(n);i++)
#define ms(x,a)    memset(x,a,sizeof(x))
#define all(x)     (x).begin(),(x).end()
#define sz(x)      int(x.size())
#define pb         push_back
#define sc(x)      scanf("%d",&x)
#define scl(x)     scanf("%lld",&x)
#define pr(x)      printf("%d ",x)
#define prl(x)     printf("%lld ",x)
#define prln(x)    printf("%d\n",x)
#define prlln(x)   printf("%lld\n",x)
#define r0         return 0;
#define CC         continue;
#define scd(x)     scanf("%lf",&x)
#define prd(x)     printf("%.10f\n",x)
#define bb(x)      cerr<< x <<endl;
#define pii        pair<int,int>
#define pis        pair<int,string>
#define vi         vector<int>
#define xx          first
#define yy          second
//#define max(a,b) a<b?b:a  __builtin_popcount;
//#define min(a,b) a<b?a:b  freopen("output.txt","w",stdout),freopen("input.txt","r",stdin);
const ll M  = 1e9+7 ;
const int N = 50500;
const int X = 0x7f7f7f7f;
const double eps = 1e-8;
//////////////////////    Hi    ^_^  hack me if you can :P
int n,m;
int t[101];
int t2[101];
int p[101];
int x[101];
ll ans = 1e18;
ll tmp;
int main()
{
    cin>>n;
    tmp = 0LL;
    lop(i,n){
        cin>>t[i];
        tmp+=1LL*t[i];
    }
    if(tmp<ans)
        ans = tmp;
    cin>>m;
    lop(i,m){
        cin>>p[i];
        p[i]--;
        cin>>x[i];
        cout<<ans-t[p[i]]+x[i]<<endl;
    }

    r0;
}
/*
2 2
..
.X

, vector < pair <int,int > >, greater <pair <int, int> >
http://codeforces.com/contest/295/submission/17929453
freopen("output.txt","w",stdout);
freopen("input.txt","r",stdin);

ll pwmod(ll a, ll n, ll mod)
{
    ll ret = 1;
    while (n)
    {
        if (n & 1) ret = ret * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ret;
}
Wj0944296025*
http://ideone.com/KrXO2S

for (int i=0; i<n; i++)            Hi    ^_^  hack me if you can :
    P
int mx[]= {0,0,1,-1, 1,1,-1,-1};
int my[]= {1,-1,0,0, 1,-1,1,-1};
bool inside (int i,int j)
{
    return (i>=0 && i<n && j>=0 && j<m);
}

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void gcd(LL n,LL m,LL &d,LL &xx,LL &yy)
{
    if(!m)
    {
        d=n;
        xx=1;
        yy=0;
    }
    else
    {
        gcd(m,n%m,d,yy,xx);
        yy-=xx*(n/m);
    }
}

int getm(int l,int r)
{
    if(l>r)
        return N;
    if(l==r)
        return 0;
    if(l==r-1)
        return s[l] != s[r];
    if(s[l]==s[r])
        return getm(l+1,r-1);
    else
        return 1+min(getm(l+1,r),getm(l,r-1));
}

int BIT[si], a[si], n;
void update(int x, int delta)
{
    for(; x <= N; x += x&-x)
        BIT[x] += delta;
}
int query(int x)
{
    int sum = 0;
    for(; x > 0; x -= x&-x)
        sum += BIT[x];
    return sum;
}

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order

*/