## Coordinate Compression: Range Sum

Suppose there is an infinite number line, and the number on each coordinate on the number line is 0.

Now we first proceed n operations, each operation will be add c on a certain position x.

Next, proceed m queries, each query contains two integers l with r, You need to find for the interval [l, r], the sum of all the numbers between index l and index r.

Input format The first line contains two integers n with m.

Next n rows, each row contains two integers x and c.

Next m rows, each row contains two integers l and r.

Output format Total m Line, each line outputs the sum of the numbers in the interval sought in the query.

data range −109≤ x≤109, 1 ≤ n , m ≤105, −109≤ l≤ r≤109, − 10000 ≤ c ≤ 10000

Sample Input:

```
3 3
1 2
3 6
7 5
1 3
4 6
7 8
```

Sample Output:

```
8
0
5
```

### Solution

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#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> pi;
const int N = 300010;
int a[N], s[N];
int n, m;
vector<int> alls;
vector<pi> add, query;
int find(int x)
{
/*
int l = 0, r = alls.size() - 1;
while(l < r)
{
int mid = l + r >> 1;
if(alls[mid] >= x) r = mid;
else l = mid + 1;
}
return r + 1;
*/
return upper_bound(alls.begin(), alls.end(), x) - alls.begin();
}
int main()
{
cin >> n >> m;
for(int i = 0; i < n; i ++ )
{
int x, c;
cin >> x >> c;
add.push_back({x, c});
alls.push_back(x);
}
for(int i = 0; i < m; i ++ )
{
int l, r;
cin >> l >> r;
query.push_back({l, r});
alls.push_back(l);
alls.push_back(r);
}
sort(alls.begin(), alls.end());
alls.erase( unique( alls.begin(), alls.end() ), alls.end());
for(auto item : add)
{
int x = find(item.first);
a[x] += item.second;
}
for(int i = 1; i <= alls.size(); i ++ ) s[i] = s[i - 1] + a[i];
for(auto item : query)
{
int l = find(item.first), r = find(item.second);
cout << s[r] - s[l - 1] << endl;
}
return 0;
}