codeforces379C

New Year Ratings Change

 CodeForces - 379C 

One very well-known internet resource site (let's call it X) has come up with a New Year adventure. Specifically, they decided to give ratings to all visitors.

There are n users on the site, for each user we know the rating value he wants to get as a New Year Present. We know that user i wants to get at least a_i rating units as a present.

The X site is administered by very creative and thrifty people. On the one hand, they want to give distinct ratings and on the other hand, the total sum of the ratings in the present must be as small as possible.

Help site X cope with the challenging task of rating distribution. Find the optimal distribution.

Input

The first line contains integer n (1 ≤ n ≤ 3·10⁵) — the number of users on the site. The next line contains integer sequence a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

Print a sequence of integers b₁, b₂, ..., b_n. Number b_i means that user i gets b_i of rating as a present. The printed sequence must meet the problem conditions.

If there are multiple optimal solutions, print any of them.

Examples

Input

3
5 1 1

Output

5 1 2

Input

1
1000000000

Output

1000000000

sol：XJB乱构造即可，每次要么是ai，否则就是上一个加1

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=300005;
int n,ans[N];
struct Record
{
    int Num,Id;
    inline bool operator<(const Record &tmp)const
    {
        return Num<tmp.Num;
    }
}a[N];
int main()
{
    int i,Now=1;
    R(n);
    for(i=1;i<=n;i++) R(a[a[i].Id=i].Num);
    sort(a+1,a+n+1);
    for(i=1;i<=n;i++)
    {
        ans[a[i].Id]=max(Now,a[i].Num);
        Now=max(Now+1,a[i].Num+1);
    }
    for(i=1;i<=n;i++) W(ans[i]);
    return 0;
}
/*
Input
3
5 1 1
Output
5 1 2

Input
1
1000000000
Output
1000000000
*/