題目連結 ：http://codeforces.com/gym/101466/problem/A

A. Gaby And Addition

time limit per test

6.0 s

memory limit per test

1024 MB

input

standard input

output

standard output

Gaby is a little baby who loves playing with numbers. Recently she has learned how to add 2 numbers using the standard addition algorithm which we summarize in 3 steps:

1. Line up the numbers vertically matching digits places.
2. Add together the numbers that share the same place value from right to left.
3. Carry if necessary.

it means when adding two numbers we will get something like this:

Unfortunately as Gaby is too young she doesn't know what the third step means so she just omitted this step using her own standard algorithm (Gaby's addition algorithm). When adding two numbers without carrying when necessary she gets something like the following:

Gaby loves playing with numbers so she wants to practice the algorithm she has just learned (in the way she learned it) with a list of numbers adding every possible pair looking for the pair which generates the largest value and the smallest one.

She needs to check if she is doing it correctly so she asks for your help to find the largest and the smallest value generated from the list of numbers using Gaby's addition algorithm.

Input

The input starts with an integer n (2 ≤ n ≤ 106) indicating the number of integers Gaby will be playing with. The next line contains nnumbers ni (0 ≤ ni ≤ 1018) separated by a single space.

Output

Output the smallest and the largest number you can get from adding two numbers from the list using Gaby's addition algorithm.

Examples

input

Copy

6
17 5 11 0 42 99

output

Copy

0 99

input

Copy

7
506823119072235413 991096248449924896 204242310783332529 778958050378192979 384042493592684633 942496553147499866 410043616343857825

output

Copy

52990443860776502 972190360051424498

Note

In the first sample input this is how you get the minimum and the maximum value

題意：給 n個數求不進位加法，兩個數和的最大值，最小值。

題解：分別對每個數字拆分成 18個位，每個位是 0-9 的數字，然後用每個位建一個字典樹。

就形成了一棵以0結點為根節點，然後每層分配0-9 兒子節點的字典樹，然後每次查詢和當前值相加最大值和最小值，分別每次從取模10最大的和最小的節點匹配。

例如 當前位是    5 最大值直接從當前位兒子節點  4開始找 如果存在直接求兩個數的和，否則繼續3 2 1.。。。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;

#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));

const long long mod=1e9+7;
const int maxn=1e6+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int n;
int cnt=1;
struct Trie {
    int son[10];
    void init() {
        for(int i=0; i<10; i++) {
            son[i]=-1;
        }
    }
} tree[maxn*20];
ll p[20];
void insert(int r,int pos,ll val) {
    if(pos==-1)return ;
    ll v=val/p[pos]%10;
    if(tree[r].son[v]==-1) {
//        printf("%d",v);
        tree[cnt].init();
        tree[r].son[v]=cnt++;
    }
    insert(tree[r].son[v],pos-1,val);
}

ll findmx(int r,int pos,ll val) {
    if(pos==-1)return 0;
    ll v=val/p[pos]%10;
    for(int i=9-v; i>=0; i--) {
        if(tree[r].son[i]!=-1) {
            return (p[pos]*((v+i)%10))+findmx(tree[r].son[i],pos-1,val);
        }
    }
    for(int i=9; i>9-v; i--)
        if(tree[r].son[i]!=-1)
            return (p[pos]*((v+i)%10))+findmx(tree[r].son[i],pos-1,val);
}
ll findmi(int r,int pos,ll val) {
    if(pos==-1)return 0;
    int v=val/p[pos]%10;
    for(int i=10-v; i<=9; i++)
        if(tree[r].son[i]!=-1) {
            return (p[pos]*((v+i)%10))+findmi(tree[r].son[i],pos-1,val);
        }
    for(int i=0; i<10-v; i++)
        if(tree[r].son[i]!=-1)
            return (p[pos]*((v+i)%10))+findmi(tree[r].son[i],pos-1,val);
}
int main() {
    scanf("%d",&n);
    p[0]=1;
    for(int i=1; i<=18; i++) {
        p[i]=p[i-1]*10;
    }
    tree[0].init();
    ll mx=-1e18,mi=1e18;
    for(int i=0; i<n; i++) {
        ll x;
        scanf("%lld",&x);
        if(i!=0) {
            mi=min(mi,findmi(0,18,x));
            mx=max(mx,findmx(0,18,x));
        }
        insert(0,18,x);
//        puts("");
    }
    printf("%lld %lld\n",mi,mx);
    return 0;
}