Codeforces Good Bye 2016 C.New Year and Rating//qduoj 三億河老師的上分夢想

C. New Year and Rating time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Every Codeforces user has rating, described with one integer, possibly negative or zero. Users are divided into two divisions. The first division is for users with rating 1900 or higher. Those with rating 1899 or lower belong to the second division. In every contest, according to one's performance, his or her rating changes by some value, possibly negative or zero.

Limak competed in n contests in the year 2016. He remembers that in the i-th contest he competed in the division di (i.e. he belonged to this division just before the start of this contest) and his rating changed by ci just after the contest. Note that negative ci denotes the loss of rating.

What is the maximum possible rating Limak can have right now, after all n contests? If his rating may be arbitrarily big, print "Infinity". If there is no scenario matching the given information, print "Impossible".

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000).

The i-th of next n lines contains two integers ci and di ( - 100 ≤ ci ≤ 100, 1 ≤ di ≤ 2), describing Limak's rating change after the i-th contest and his division during the i-th contest contest.

Output

If Limak's current rating can be arbitrarily big, print "Infinity" (without quotes). If the situation is impossible, print "Impossible" (without quotes). Otherwise print one integer, denoting the maximum possible value of Limak's current rating, i.e. rating after the ncontests.

Examples input

3
-7 1
5 2
8 2
      

output

1907
      

input

2
57 1
22 2
      

output

Impossible
      

input

1
-5 1
      

output

Infinity
      

input

4
27 2
13 1
-50 1
8 2
      

output

1897
      

Note

In the first sample, the following scenario matches all information Limak remembers and has maximum possible final rating:

• Limak has rating 1901 and belongs to the division 1 in the first contest. His rating decreases by 7.
• With rating 1894 Limak is in the division 2. His rating increases by 5.
• Limak has rating 1899 and is still in the division 2. In the last contest of the year he gets  + 8 and ends the year with rating 1907.

In the second sample, it's impossible that Limak is in the division 1, his rating increases by 57 and after that Limak is in the division 2 in the second contest.

根據下場比賽的等級和本場比賽rank的變化（其實也就是根據本場比賽的rank值變化，和本場比賽後的等級）确定一個區間，還有通過之前比賽得到的一個區間，判斷一下，每次修改一下區間

#include<stdio.h>
#include<queue>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int c[200005],d[200005];
int main()
{
	int le,ri,tle,tri,n,i,flog=0;
	while(~scanf("%d",&n))
	{
		flog=0;
		for(i=0;i<n;i++)
	scanf("%d %d",&c[i],&d[i]);
	if(d[0]&1)
	{
		le=1900;
		ri=inf;
	}
	else
	{
		le=-1*inf;
		ri=1899;
	}
	for(i=0;i<n-1;i++)
	{
		if(flog)
		break;
		if(d[i+1]&1)
		{
			tle=1900-c[i];
			tri=inf;
			if(ri<tle||le>tri)
			{
				flog=1;break;
			}
			ri=min(ri,tri);
			le=max(le,tle);
		}
		else
		{
			tri=1899-c[i];
			tle=-1*inf;
			if(ri<tle||le>tri)
			{
				flog=1;break;
			}
			ri=min(ri,tri);
			le=max(le,tle);
		}
		if(le!=-1*inf)
		le+=c[i];
		if(ri!=inf)
		ri+=c[i];
	}
	if(flog)
	printf("Impossible\n");
	else if(ri>=inf)
	printf("Infinity\n");
	else
	printf("%d\n",ri+c[n-1]);
	}
	
	return 0;
}