/*
problem: FatMouse' Trade
this is greedy problem.
firstly:we should calculate the average J[i]/F[i] and sort it
secondly: according to the m and choose the most high ones
*/
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct cell
{
double j;
double f;
double rate;
};
bool cmp(cell x,cell y)
{
return x.rate > y.rate;
}
int main()
{
int m,n;
cell temp;
vector<cell> vec;
while (scanf("%d%d", &m, &n) != EOF)
{
if ((m == -1)&&(n == -1))
break;
double count = 0.0;
while (n--)
{
cin>>temp.j>>temp.f;
if (temp.f != 0)
{
temp.rate = (double)temp.j/temp.f;
vec.push_back(temp);
}
else
{
count += temp.j;
}
}
sort(vec.begin(), vec.end(), cmp);
for (int i = 0; (i < vec.size())&&(m > 0); i++)
{
if (m-vec[i].f >= 0)
{
count += vec[i].j;
m -= vec[i].f;
}
else
{
//count += (double)((double)(m/vec[i].f))*vec[i].j;
double temp1 = (double)m/vec[i].f;
count += temp1*vec[i].j;
m = 0;
}
}
vec.clear();
printf("%.3f\n",count);
}
}
/*此题除了要满足例子以外，还要满足一些条件才能真正算ac：
0 1
1 0
1.000
1 0
0.000
5 4
10000 5
2000 2
100 0
300 0
10400.000
数据类型用double,就这样*/
