Codeforces Round #347 (Div. 2) B. Rebus

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 from
1 to n, such that equality holds.

Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than
100 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 from
1 to n. Follow the format given in the samples.

Examples

Input

? + ? - ? + ? + ? = 42

Output

Possible
9 + 13 - 39 + 28 + 31 = 42

Input

? - ? = 1

Output

Impossible

Input

? = 1000000

Output

Possible
1000000 = 1000000

分析：设加号有 a - 1个，减号有b个，那么当且仅当a - b*n <= n <= a*n - b 时有解，解可以二分出加减的总值后构造出。

#include <cstdio>
#include <iostream>
using namespace std;
string S;
int n,tot,a,b,f[105],A[105],B[105];
int main()
{
a++;
while(cin>>S && S!="=")
{
if(S == "?") continue;
if(S == "+") a++,f[++tot] = 1;
else b++,f[++tot] = -1;
}
cin>>n;
if(a*n - b < n || a - b*n > n)
{
cout<<"Impossible"<<endl;
return 0;
}
else cout<<"Possible"<<endl;
int s = b,t = b*n;
while(s != t)
{
int mid = (s + t) / 2;
if(mid + n >= a) t = mid;
else s = mid + 1;
}
s -= b;
for(int i = 1;i <= b;i++)
{
B[i] = 1;
B[i] += min(n-1,s);
s -= min(n-1,s);
}
t += n - a;
for(int i = 1;i <= a;i++)
{
A[i] = 1;
A[i] += min(n-1,t);
t -= min(n-1,t);
}
cout<<A[1];
int nowa = 2,nowb = 1;
for(int i = 1;i <= tot;i++)
if(f[i] == 1) cout<<" + "<<A[nowa++];
else cout<<" - "<<B[nowb++];
cout<<" = "<<n<<endl;
}