## KGSS - Maximum Sum

You are given a sequence A[1], A[2], ..., A[N] ( 0 ≤ A[i] ≤ 10^8 , 2 ≤ N ≤ 10^5 ). There are two types of operations and they are defined as follows:

Update:

This will be indicated in the input by a 'U' followed by space and then two integers i and x.

U i x, 1 ≤ i ≤ N, and x, 0 ≤ x ≤ 10^8.

This operation sets the value of A[i] to x.

Query:

This will be indicated in the input by a 'Q' followed by a single space and then two integers i and j.

Q x y, 1 ≤ x < y ≤ N.

You must find i and j such that x ≤ i, j ≤ y and i != j, such that the sum A[i]+A[j] is maximized. Print the sum A[i]+A[j].

### Input

The first line of input consists of an integer N representing the length of the sequence. Next line consists of N space separated integers A[i]. Next line contains an integer Q, Q ≤ 10^5, representing the number of operations. Next Q lines contain the operations.

### Output

Output the maximum sum mentioned above, in a separate line, for each Query.

Input:
5
1 2 3 4 5
6
Q 2 4
Q 2 5
U 1 6
Q 1 5
U 1 7
Q 1 5

Output:
7
9
11
12

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

#define MOD                 1000000007LL
#define EPS                 1e-9
#define io                  ios_base::sync_with_stdio(false);cin.tie(NULL);

const int MAXN = 1e5+5;
int arr[MAXN];
struct Tree{int largest, second_largest;};
Tree T[4*MAXN];
int n;

void build(int node, int start, int end){
if(start > end)
return;
if(start == end){
T[node].largest = arr[start];
T[node].second_largest = 0;
return;
}
int left = node << 1, right = left + 1;
int mid = (start + end) >> 1;
build(left, start, mid);
build(right, mid + 1, end);
if(T[left].largest >= T[right].largest){
T[node].largest = T[left].largest;
T[node].second_largest = max(T[right].largest, T[left].second_largest);
}else{
T[node].largest = T[right].largest;
T[node].second_largest = max(T[left].largest, T[right].second_largest);
}
}

Tree query(int node, int start, int end, int l, int r){
if(start > end || start > r || end < l){
Tree t;
t.largest = t.second_largest = 0;
return t;
}
if(start >= l && end <= r)
return T[node];
int left = node << 1, right = left + 1;
int mid = (start + end) >> 1;
Tree t1 = query(left, start, mid, l, r);
Tree t2 = query(right, mid + 1, end, l, r);
Tree t3;
if(t1.largest >= t2.largest){
t3.largest = t1.largest;
t3.second_largest = max(t2.largest, t1.second_largest);
}else{
t3.largest = t2.largest;
t3.second_largest = max(t2.second_largest, t1.largest);
}
return t3;
}

void update(int node, int start, int end, int idx, int val){
if(start > end)
return;
if(start == end){
arr[start] = val;
T[node].largest = val;
T[node].second_largest = 0;
return;
}
int left = node << 1, right = left + 1;
int mid = (start + end) >> 1;
if(idx <= mid)
update(left, start, mid, idx, val);
else
update(right, mid + 1, end, idx, val);
if(T[left].largest >= T[right].largest){
T[node].largest = T[left].largest;
T[node].second_largest = max(T[right].largest, T[left].second_largest);
}else{
T[node].largest = T[right].largest;
T[node].second_largest = max(T[left].largest, T[right].second_largest);
}
}

int main(){
io;
scanf("%d", &n);
for(int i = 0;i < n; i++)
scanf("%d", arr+i);
build(1, 0, n-1);
int q;
scanf("%d ", &q);
while(q--){
char t;
int l, r;
scanf("%c%d%d ", &t, &l, &r);
if(t == 'U'){
l--;
update(1, 0, n-1, l, r);
}else{
l--;r--;
Tree t = query(1, 0, n-1, l, r);
printf("%d\n", t.largest + t.second_largest);
}
}
return 0;
}
Copy The Code &

Input

cmd
5
1 2 3 4 5
6
Q 2 4
Q 2 5
U 1 6
Q 1 5
U 1 7
Q 1 5

Output

cmd
7
9
11
12
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## Demonstration

SPOJ Solution-Maximum Sum-Solution in C, C++, Java, Python