GoodBye 2016 C. New Year and Rating 详解（维护不等式）

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.

题目大意：

每个人都有一个rating，是一个整数，可以是负数以及0，div1是>=1900分 div2是<=1899，然后有个人参加了N场，比赛，但是只记得每一场是div几，以及每一场赛后的rating的变化量，请问他现在最高可能多少分，如果不可能有符合的情况
输出Impossible，如果可以无限大分数，输出Infinity
思路：首先，明确一下题目问的是一顿比赛之后，现在的成绩最高可能是多少，不是过程中的，比赛一直当做过程中做的。。。假设x为初始分数，presum是这场之前所有变化的和，如果这场结束后是div1，那么x+presum
>= 1900符合题意，如果这场结束之后，是div2，那么x+presum <= 1899，综合一下就是  1900- presum <= x <= 1899- presum x满足这个不等式，要对所有比赛都成立，这就要求，后面的presum最大，这样每场x都就满足1899 - presum了， 前面的最小，这样就满足 所有x 都 》= 1900 - presum了，如果全部比赛结束后， 后面presum >= 前面presum 这样等式肯定不成立了，就是impossible，如果一场div2都没有，那样x就随意大了，所以不确定，剩下的就是取初始值x最大的情况即
1899-presum，然后加上所有的变化和，明确一点，所有presum都是不变的，变得只有初始值x。

第一次做这种维护不等式的题，一开始想的是，让x=1899，然后不断循环，但是忘记了所有presum都是一定的，这样 只有初始值x是变量，在循环中维护x的最大最小值就行。。

#include <iostream>
#include <cstdio>
using namespace std;
const int INF = 2e8;
int main()
{
int n, presum, low, high, rank, a; //low，high记录让不等式成立的最大最小presum
while(~scanf("%d", &n))
{
low = INF;
high = -INF;
presum = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d%d", &a, &rank);
if(rank == 2)
high = max(presum, high);
if(rank == 1)
low = min(presum, low);
presum += a;
}
if(high == -INF)
cout << "Infinity" << endl;
else if(low <= high)
cout << "Impossible" << endl;
else
cout << 1899-high+presum << endl;
}
return 0;
}