W越大，Y越小，所以可以二分W使Y接近S。
少用min、abs可以讓思路更清晰。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn = 200010;
int n, m, num[maxn];
int mw;
long long s, sum[maxn], tot;

struct node1{
    int w, v;
} a[maxn];

struct node2{
    int
 
 li, ri;
} b[maxn];

long long judge(int x){
    memset(sum, 0, sizeof(sum));
    memset(num, 0, sizeof(num));
    for(int i = 1; i <= n; i++){
        sum[i] = sum[i-1], num[i] = num[i-1];
        if(a[i].w > x){
            sum[i] += a[i].v;
            num[i]++;
        }
    }

    tot = 0;
    for
 
(int i = 1; i <= m; i++)
        tot += (sum[b[i].ri] - sum[b[i].li-1])*(num[b[i].ri] - num[b[i].li-1]);
    return tot;
}

int main(){
    cin >> n >> m >> s;
    for(int i = 1; i <= n; i++){
        cin >> a[i].w >> a[i].v;
        mw = max(mw, a[i].w);
    }

    for
 
(int i = 1; i <= m; i++)
        cin >> b[i].li >> b[i].ri;
    long long ans = s;
    int l = 0, r = mw + 1;
    while(l < r){
        int mid = (l+r) >> 1;
        long long now = judge(mid);
        if(now > s){
            l = mid + 1;
            if(ans > now - s)   ans = now - s;
        }
        else if(now < s){
            r = mid;
            if(ans > s - now)   ans = s - now;
        }
        else{
            cout << "0" << endl;
            return 0;
        }
    }
    cout << ans << endl;
    return 0;
}