Problem:
Problem Statement

You are given two integers a and b (a≤b). Determine if the product of the integers a, a+1, …, b is positive, negative or zero.

Constraints

a and b are integers.

-10^9≤a≤b≤10^9

Partial Score

In test cases worth 100 points, -10≤a≤b≤10.

Input

The input is given from Standard Input in the following format:

a b

Output

If the product is positive, print Positive. If it is negative, print Negative. If it is zero, print Zero.

Sample Input 1

1 3

Sample Output 1

Positive

1×2×3=6 is positive.

Sample Input 2

-3 -1

Sample Output 2

Negative

(-3)×(-2)×(-1)=-6 is negative.

Sample Input 3

-1 1

Sample Output 3

Zero

(-1)×0×1=0.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

typedef long long ll;
typedef long double R;
const R EPS=1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max(x,0.0L));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

// Problem Specific Parameter:

int main() {
  ll a, b; cin >> a >> b;
  if(a > 0) cout << "Positive" << endl;
  else if(a <= 0 && b >= 0) cout << "Zero" << endl;
  else if(a < 0 && b < 0) {
    int tmp = b - a + 1;
    if(tmp % 2 == 0) cout << "Positive" << endl;
    else cout << "Negative" << endl;
  }

  return 0;
}