The Number of Products

You are given a sequence a1,a2,…,ana1,a2,…,an consisting of nn non-zero integers (i.e. ai≠0ai≠0).

You have to calculate two following values:

the number of pairs of indices (l,r)(l,r) (l≤r)(l≤r) such that al⋅al+1…ar−1⋅aral⋅al+1…ar−1⋅ar is negative;
the number of pairs of indices (l,r)(l,r) (l≤r)(l≤r) such that al⋅al+1…ar−1⋅aral⋅al+1…ar−1⋅ar is positive;

Input

The first line contains one integer nn (1≤n≤2⋅105)(1≤n≤2⋅105) — the number of elements in the sequence.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109;ai≠0)(−109≤ai≤109;ai≠0) — the elements of the sequence.

Output

Print two integers — the number of subsegments with negative product and the number of subsegments with positive product, respectively.

Examples

Input

5
5 -3 3 -1 1

Output

8 7

Input

10
4 2 -4 3 1 2 -4 3 2 3

Output

28 27

Input

5
-1 -2 -3 -4 -5

Output

9 6

题解及分析

AC代码:

#include<stdio.h>
#include<string.h>
//R[i]表示以i为结尾的正子列的个数,L[i]表示以i为结尾的负子列的个数
long long int n,a[200005],R[200005],L[200005],ansR=0,ansL=0;
int main()
{scanf("%lld",&n);for(long long int i=1;i<=n;i++){scanf("%lld",&a[i]);//将a[i]按照是否为负数分类,便于后面进行遍历if(a[i]<0) a[i]=1;else a[i]=0;}R[0]=0;L[0]=0;for(int i=1;i<=n;i++){if(a[i])//递推关系{L[i]=R[i-1]+1;R[i]=L[i-1];}else {R[i]=R[i-1]+1;L[i]=L[i-1];}ansR+=R[i];//ansR表示前i个数组成的数列一共有ansR个正子列ansL+=L[i];//ansL表示前i个数组成的数列一共有ansL个负子列}printf("%lld %lld",ansL,ansR);return 0;
}