C. New Year and Rating

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 d_i (i.e. he belonged to this division just before the start of this contest) and his rating changed by c_i just after the contest. Note that negative c_i 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 c_i and d_i ( - 100 ≤ c_i ≤ 100, 1 ≤ d_i ≤ 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 n contests.

Examples

Input

3 -7 1 5 2 8 2

Output

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

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.

/*
给出每次的上的分和降的分，和相应的等级，让你输出最后的分数，不可能输出impossible。可以是正无穷的话Infinity，否则输出最大的.

初步思路：建立两个边界，从第一条开始遍历，根据条件放缩，到最后如果最大边界不确定，就是正无穷，左边界大于有边界就是不可能，否则输出有边界

#反思：比赛的时候放缩的时候想的太复杂了。分情况的时候取大取小没考虑好
       好在有惊无险的上分了，离紫色又进了一步
*/
#include<bits/stdc++.h>
#define INF 300000000
using namespace std;
long long n;
long long a,b;
long long l,r;
int main(){
    //freopen("in.txt","r",stdin);
    cin>>n;
    l=-INF,r=INF;//判断的两个边
    for(int i=0;i<n;i++){
        cin>>a>>b;
        if(b==1&&l<1900) l=1900;//当前是第一阶段，如果本来的边界有小于1900的，那么不满足，必须放缩到1900
        if(b==2&&r>1899) r=1899;
        l+=a;
        r+=a;
    }
    long long cur=max(l,r);
    if(l>r) puts("Impossible");
    else if(cur>200000000) puts("Infinity");//就算2000000次都上分，那么最大也只可能是200000000
    else cout<<cur<<endl;
    return 0;
}