This time, you are supposed to find A*B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:K N1 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

2 1 2.4 0 3.2
2 2 1.5 1 0.5

3 3 3.6 2 6.0 1 1.6

/**
題意：
兩個多項式，係數是浮點數。求乘積
題解：
陣列 a[i] i表指數，a[i]表係數
模擬,

**注意！！最大值開到2000，1000+1000
*/

#include <cstdio>
#include <cassert>
#include <iostream>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <math.h>

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 2000 + 5;
double a[maxn];
double b[maxn];
double c[maxn];
int main() {
    int n,k;
    double t;
    scanf("%d",&n);
    for(int i = 0;i<n;++i){
        scanf("%d%lf",&k,&t);
        a[k] = t;
    }
    scanf("%d",&n);
    for(int i = 0;i<n;++i){
        scanf("%d%lf",&k,&t);
        b[k] = t;
    }
    for(int i = 0;i<maxn;++i){
        for(int j = 0;j<maxn;++j){
            if(a[i]==0 || b[j]==0) continue;
            c[i+j] += a[i]*b[j];
        }
    }
    n = 0;
    vector<int> ans;
    for(int i = maxn-1;i>=0;--i){
        if(fabs(c[i]-0)>1e-8){
           n++;
           ans.push_back(i);
        }
    }
    printf("%d",n);
    for(int i = 0;i<ans.size();++i){
        printf(" %d %.1f",ans[i],c[ans[i]]);
    }
    return 0;
}