#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 3e6 + 10;
int n, m;
int a[N], s[N];
vector<int> alls;
vector<PII> 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;
}
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;
}
//如果数范围在 10^5次方以内 就是前缀和算法
// x.c a[x] += c
// s[r] - s[l - 1] 前缀和公式
//此题讲离散化 所以用离散的方法
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 3e6 + 10;
int n, m;
int a[N], s[N];
vector<int> alls;
vector<PII> add, query;
// vector<int>::iterator unique(vector<int> &a){
// //java python实现unique函数
// int j = 0;
// for(int i = 0; i < a.size(); i++)
// if(!i || a[i] != a[i - 1])
// a[j++] = a[i];
// //a[0] ~ a[j - 1] 所有a中不重复的数
// return a.begin() + j;
// }
// java实现
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;
}
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);
}
//去重
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());
//处理插入
for(auto item : add){
int x = find(item.first);
a[x] += item.second;
}
//预处理前缀和
for(int i = 1; i <= alls.size(); i++) s[i] = s[i - 1] + a[i];
//处理询问
for(auto item : query){
int l = find(item.first), r = find(item.second);
cout << s[r] - s[l - 1] << endl;
}
return 0;
}