$Description$Description

给定一个合法的多项式，求导

数据范围：

---

$Solution$Solution

字符串处理题，简单

---

$Code$Code

#include<cstdio>
#include<cctype>
#include<cstring> 
#include<algorithm> 
using namespace std;char s[2000001]; 
struct mo{long long a,b;}a[1001];int n;
long long now;
bool flag,flag2,flag3;
inline mo qd(mo x){return (mo){x.a*x.b,x.b-1};}
signed main()
{
	freopen("equation.in","r",stdin);
	freopen("equation.out","w",stdout);
	scanf("%s",s+1);
	for(register int i=1;i<=strlen(s+1);i++)
	{
		if(isdigit(s[i]))
		{
			now=(now<<3)+(now<<1)+s[i]-48;
			continue;
		}
		if(s[i]=='-')
		{
			if(!flag) {flag3=true;now=0;}
			else
			if(flag2)
			{
				a[n].b=now;
				now=0;flag=flag2=false;flag3=true;
			}
			else
			{
				if(flag3) a[++n]=(mo){-now,0};else a[++n]=(mo){now,0};
				now=0;flag=flag2=false;flag3=true;
			}
			continue;
		}
		if(s[i]=='+')
		{
			if(flag2)
			{
				a[n].b=now;
				now=0;flag=flag2=flag3=false;
			}
			else
			{
				if(flag3) a[++n]=(mo){-now,0};else a[++n]=(mo){now,0};
				now=0;flag=flag2=flag3=false;
			}
		} 
		if(s[i]=='x')
		{
			if(!now) now=1;
			if(flag3) now*=-1;
			a[++n]=(mo){now,1};
			flag=true;now=0;
			continue;
		}
		if(s[i]=='^')
		{
			now=0;flag2=true;
			continue;
		}
	}
	if(now>0) 
		if(!flag2)
			if(!flag3)
			  a[++n]=(mo){now,0};
			else 
			  a[++n]=(mo){-now,0};
		else
			a[n].b=now;
	flag=false;
	for(register int i=1;i<=n;i++) 
	{
		a[i]=qd(a[i]);
		if(!a[i].a) continue;
		if(a[i].a>0&&flag)putchar('+');
		if(!a[i].b) {printf("%lld",a[i].a);flag=true;continue;}
		if(abs(a[i].a)>1) printf("%lld",a[i].a);if(a[i].a==-1) putchar('-');putchar('x');
		if(a[i].b>1) printf("^%lld",a[i].b);
		flag=true;
	}
	if(!flag) putchar(48);
}