codeforces-747

题目链接：点击打开链接

A. Display Size

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly n pixels.

Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels a and the number of columns of pixelsb, so that:

• there are exactly n pixels on the display;
• the number of rows does not exceed the number of columns, it means a ≤ b;
• the difference b - a is as small as possible.

Input

The first line contains the positive integer n (1 ≤ n ≤ 106) — the number of pixels display should have.

Output

Print two integers — the number of rows and columns on the display.

Examples

input

8

output

2 4

input

64

output

8 8

input

5

output

1 5

input

999999

output

999 1001

Note

In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.

In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.

In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.

#include<cstdio>
#include<cmath>
using namespace std;
int n;
int main()
{
	while(~scanf("%d",&n))
	{
		int mid=sqrt(n*1.0);
		int ans;
		for(int i=mid;i>=1;i--)
		{
			if(n%i==0)
			{
				ans=i;
				break;
			}
		}
		printf("%d %d\n",ans,n/ans);
	}
	return 0;
}

题目链接： 点击打开链接

B. Mammoth's Genome Decoding

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The process of mammoth's genome decoding in Berland comes to its end!

One of the few remaining tasks is to restore unrecognized nucleotides in a found chain s. Each nucleotide is coded with a capital letter of English alphabet: 'A', 'C', 'G' or 'T'. Unrecognized nucleotides are coded by a question mark '?'. Thus, s is a string consisting of letters 'A', 'C', 'G', 'T' and characters '?'.

It is known that the number of nucleotides of each of the four types in the decoded genome of mammoth in Berland should be equal.

Your task is to decode the genome and replace each unrecognized nucleotide with one of the four types so that the number of nucleotides of each of the four types becomes equal.

Input

The first line contains the integer n (4 ≤ n ≤ 255) — the length of the genome.

The second line contains the string s of length n — the coded genome. It consists of characters 'A', 'C', 'G', 'T' and '?'.

Output

If it is possible to decode the genome, print it. If there are multiple answer, print any of them. If it is not possible, print three equals signs in a row: "===" (without quotes).

Examples

input

8
AG?C??CT

output

AGACGTCT

input

4
AGCT

output

AGCT

input

6
????G?

output

===

input

4
AA??

output

===

Note

In the first example you can replace the first question mark with the letter 'A', the second question mark with the letter 'G', the third question mark with the letter 'T', then each nucleotide in the genome would be presented twice.

In the second example the genome is already decoded correctly and each nucleotide is exactly once in it.

In the third and the fourth examples it is impossible to decode the genom.

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
char str[260];
int num[4];
char ch[]={'A','C','G','T'};
int main()
{
	while(~scanf("%d",&n))
	{
		scanf("%s",str);
		if(n%4)
		{
			puts("===");
			continue;
		}
		memset(num,0,sizeof(num));
		for(int i=0;i<n;i++)
		{
			if(str[i]=='A')
				num[0]++;
			if(str[i]=='C')
				num[1]++;
			if(str[i]=='G')
				num[2]++;
			if(str[i]=='T')
				num[3]++;
		}
		int cnt=n/4;
		if(num[0]>cnt||num[1]>cnt||num[2]>cnt||num[3]>cnt)
		{
			puts("===");
			continue;
		}
		for(int i=0;i<n;i++)
		{
			if(str[i]=='?')
			{
				int k;
				for(int j=0;j<4;j++)
				{
					if(num[j]<cnt)
					{
						k=j;
						num[j]++;
						break;
					}
				}
				str[i]=ch[k];
			}
		}
		puts(str);
	}
	return 0;
}