。。。。此次Div2就做了兩個題目，第一個題目wa了4次，22分鐘出的，第二道題目一遍過，18分鐘，然後看到c題感覺樣例好玄學，就搗鼓浮點數去了，沒想到是二元一次方程求解。。。還是太蠢了。

A：

You are given a string s consisting of n lowercase Latin letters.

You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.

String s=s1s2…sn

is lexicographically smaller than string t=t1t2…tm if n<m and s1=t1,s2=t2,…,sn=tn or there exists a number p such that p≤min(n,m) and s1=t1,s2=t2,…,sp−1=tp−1 and sp<tp

.

For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".

Input

The first line of the input contains one integer n

(2≤n≤2⋅105) — the length of s

.

The second line of the input contains exactly n

lowercase Latin letters — the string s

.

Output

Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string s

.

Examples

Input

Copy

3
aaa

Output

Copy

aa

Input

Copy

5
abcda

Output

Copy

abca

Note

In the first example you can remove any character of s

to obtain the string "aa".

In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".

 找第i項比第i+1項大的。刪除的就是那個字元，如果沒有，則刪除最後一位。

程式碼如下：
 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=2*1e5+5;
int n;
char a[maxn];
int main()
{
    scanf("%d",&n);
    int loc;
    char Max;
    scanf("%s",a);
    int flag=0;
    for (int i=0;i<n-1;i++)
    {
        if(a[i]>a[i+1])
        {
            flag=1;
            loc=i;
            break;
        }
    }
    if(!flag)
        loc=n-1;
    for (int i=0;i<n;i++)
        if(i!=loc)
        printf("%c",a[i]);
    printf("\n");
    return 0;
}

B. Divisor Subtraction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer number n

. The following algorithm is applied to it:

1. if n=0， then end algorithm;
2. find the smallest prime divisor d of n;
3. subtract d from n and go to step 1

• Determine the number of subtrations the algorithm will make.

  Input

  The only line contains a single integer n

  (2≤n≤1010

  ).

  Output

  Print a single integer — the number of subtractions the algorithm will make.

  Examples

  Input

  Copy

  5
  

  Output

  Copy

  1
  

  Input

  Copy

  4
  

  Output

  Copy

  2
  

  Note

  In the first example 5

  is the smallest prime divisor, thus it gets subtracted right away to make a 0

  .

  In the second example 2

  is the smallest prime divisor at both steps.

當為偶數時，直接/2即可，當為奇數，求出列舉求出最小素因子，然後減去這個素因子，必定為偶數，然後轉化為偶數求解的方式。

程式碼如下：
 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <math.h>
using namespace std;
typedef long long ll;
ll n;
int is_su (ll x,ll& yin)
{
    ll m=(ll)sqrt(x);
    for (ll i=2;i<=m;i++)
    {
        if(x%i==0)
        {
            yin=i;
            return 0;
        }
    }
    return 1;
}
int main()
{
    scanf("%lld",&n);
    if(n%2==0)
    {
        printf("%lld\n",n/2);
    }
    else
    {
        ll t;
        if(is_su(n,t))
        {
            printf("1\n");
        }
        else
        {
            printf("%lld\n",(n-t)/2+1);
        }
    }
    return 0;
}

C. Meme Problem

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Try guessing the statement from this picture:

You are given a non-negative integer d

. You have to find two non-negative real numbers a and b such that a+b=d and a⋅b=d

.

Input

The first line contains t

(1≤t≤103

) — the number of test cases.

Each test case contains one integer d

(0≤d≤103)

.

Output

For each test print one line.

If there is an answer for the i

-th test, print "Y", and then the numbers a and b

.

If there is no answer for the i

-th test, print "N".

Your answer will be considered correct if |(a+b)−a⋅b|≤10−6

and |(a+b)−d|≤10−6

.

Example

Input

Copy

7
69
0
1
4
5
999
1000

Output

Copy

Y 67.985071301 1.014928699
Y 0.000000000 0.000000000
N
Y 2.000000000 2.000000000
Y 3.618033989 1.381966011
Y 997.998996990 1.001003010
Y 998.998997995 1.001002005

 這題目想了50分鐘結果想歪了。。。

就是個解二元一次方程組的問題。

a+b=d; a*b=d;

則a*(d-a)=d;

則a*a-ad+d=0,

用求根公式求出一個解，然後d減去那個解就是b的答案。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <math.h>
using namespace std;
int t;
double d;
int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%lf",&d);
        double m;
        m=d*d-4*d;
        if(m<0)
        {
            printf("N\n");
        }
        else
        {
            double x1,x2;
            m=sqrt(m);
            x1=(d+m)/2;
            x2=d-x1;
            printf("Y %.9lf %.9lf\n",x1,x2);
        }
    }
    return 0;
}