M today, the company has a task to complete, the first task need 11 minutes to complete, at the same time, the difficulty of the task is yi, the level is lower than this level of yi machine can't complete the task, if the company to complete this task, they will get (xi yi + 2) $500,

The company has n machine, every machine has a maximum working hours and a level, if a task time exceeding the maximum working hours of the computer, the computer will not be able to complete this task, each machine a day can complete a task, each task can only be done by a machine,

The company hopes to maximize the number they can complete the task of today, if there are multiple solutions, they want to make money, to the greatest extent

Input format:
The input contains several test cases,

The first line contains two integers N and M.N is machine number. M is the number of tasks (1 & lt;=N & lt;=100000, 1 & lt;=M & lt;=100000),

The following N lines each line contains two integer xi (0 & lt; Xi & lt; 1440), yi (0=& lt; Yi & lt;=100). Xi is the maximum time machine can work. Yi is machine level,

The next M lines each line contains two integer xi (0 & lt; Xi & lt; 1440), yi (0=& lt; Yi & lt;=100). Xi is the time needed to complete the task. Yi is a task level,

The output format:
For each test case, the output two integers, the company to the maximum number of tasks can be finished today and will receive the money,

# include & lt; Iostream>
# include

using namespace std;

Struct node
{
Int x, y;
} [100005] s1 and s2 [100005].

Int CMP (node a, node b)
{
If (a.x==b.x)
Return a.y & gt; B.y.
Return a.x & gt; B.x;
}

Int main () {
Int n, m, I, j, num, l, q, t, p;
Long long money;
Cin> N> m;

for(i=0; iThe scanf (" % d % d ", & amp; S1 [I]. X, & amp; S1 [I] y);
}

for(i=0; iThe scanf (" % d % d ", & amp; [I]. S2 x, & amp; S2 [I] y);
}

Sort (s1, s1 + n, CMP);
Sort (s2, s2 + m, CMP);

Money=0;

For (I=0; iFor (j=0; jIf (s2 [I] xT=s1 [j]. Journal of x;
P=s1 [j]. J y;
Q=j;
}
}
If (t> 0 {
[q]. S1 x=0;
[q]. S1 y=0;
t=0;
P=0;
Money=500 * s2 [I]. X + 2 * s2 [I] y + money;
num++;
}
}
cout}

Why does a test by not,