Codeforces 664 B Rebus【思维】

B. Rebus

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from1 to n, such that equality holds.

Input

The only line of the input contains a rebus. It's guaranteed that it contains no more than100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.

Output

The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.

If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from1 to n. Follow the format given in the samples.

Examples

Input

? + ? - ? + ? + ? = 42

Output

Possible9 + 13 - 39 + 28 + 31 = 42

Input

? - ? = 1

Output

Impossible

Input

? = 1000000

Output

Possible1000000 = 1000000

题目大意：

在？处填入1-n的数字使得等式成立，如果不成立，输出Impossible。

思路：

先将所有？置为1.然后对应每个数进行调整。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;char a[1050];int num[1050];int abs(int aaa){    if(aaa<0)return -aaa;    else return aaa;}int main(){    while(gets(a))    {        int cont=0;        int d=1;        int val=0;        int n=strlen(a);        for(int i=0;i<n;i++)        {            if(a[i]=='+')d=1;            if(a[i]=='-')d=-1;            if(a[i]=='?')            {                num[cont++]=d;                val+=d;            }        }        int contz=1;        int sum=0;        for(int i=n-1;i>=0;i--)        {            if(a[i]>='0'&&a[i]<='9')            {                sum+=(a[i]-'0')*contz;                contz*=10;            }            else break;        }        for(int i=0;i<cont;i++)        {            while(val<sum&&num[i]>0&&num[i]<sum)num[i]++,val++;            while(val>sum&&num[i]<0&&num[i]>-sum)num[i]--,val--;        }        if(val!=sum)        {            printf("Impossible\n");            continue;        }        printf("Possible\n");        printf("%d",num[0]);        contz=1;        for(int i=1;i<n;i++)        {            if(a[i]!='?')printf("%c",a[i]);            else            {                printf("%d",abs(num[contz++]));            }        }        printf("\n");    }}