codeforces405D

Toy Sum

CodeForces - 405D

Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris.

There are exactly s blocks in Chris's set, each block has a unique number from 1 to s. Chris's teacher picks a subset of blocks X and keeps it to himself. He will give them back only if Chris can pick such a non-empty subset Y from the remaining blocks, that the equality holds:

"Are you kidding me?", asks Chris.

For example, consider a case where s = 8 and Chris's teacher took the blocks with numbers 1, 4 and 5. One way for Chris to choose a set is to pick the blocks with numbers 3 and 6, see figure. Then the required sums would be equal: (1 - 1) + (4 - 1) + (5 - 1) = (8 - 3) + (8 - 6) = 7.

However, now Chris has exactly s = 10⁶ blocks. Given the set X of blocks his teacher chooses, help Chris to find the required set Y!

Input

The first line of input contains a single integer n (1 ≤ n ≤ 5·10⁵), the number of blocks in the set X. The next line contains n distinct space-separated integers x₁, x₂, ..., x_n (1 ≤ x_i ≤ 10⁶), the numbers of the blocks in X.

Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

Output

In the first line of output print a single integer m (1 ≤ m ≤ 10⁶ - n), the number of blocks in the set Y. In the next line output m distinct space-separated integers y₁, y₂, ..., y_m (1 ≤ y_i ≤ 10⁶), such that the required equality holds. The sets X and Yshould not intersect, i.e. x_i ≠ y_j for all i, j (1 ≤ i ≤ n; 1 ≤ j ≤ m). It is guaranteed that at least one solution always exists. If there are multiple solutions, output any of them.

Examples

Input

3
1 4 5

Output

2
999993 1000000

Input

1
1

Output

1
1000000 

sol：显然可以从两边往中间做，如果一边有就弄另一边，两边都有就弄一对新的即可

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=; bool f=; char ch=' ';
    while(!isdigit(ch)) {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<)+(s<<)+(ch^); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<) {putchar('-'); x=-x;}
    if(x<) {putchar(x+''); return;}
    write(x/); putchar((x%)+'');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int S=,n,m=,a[N],b[N];
bool Bo[N];
int main()
{
    int i;
    R(n);
    for(i=;i<=n;i++) Bo[a[i]=read()]=;
    int up=S>>,cnt=;
    for(i=;i<=up;i++)
    {
        int o1=i,o2=S-i+;
        if((Bo[o1])&&(!Bo[o2]))
        {
            b[++m]=o2; Bo[o2]=;
        }
        else if((!Bo[o1])&&(Bo[o2]))
        {
            b[++m]=o1; Bo[o1]=;
        }
        else if(Bo[o1]&&Bo[o2]) cnt++;
    }
    for(i=;i<=up&&cnt;i++)
    {
        if(!Bo[i])
        {
            b[++m]=i; b[++m]=S-i+; cnt--;
        }
    }
    sort(b+,b+m+);
    Wl(m);
    for(i=;i<=m;i++) W(b[i]);
    return ;
}