#include <bits/stdc++.h>
using namespace std;
#define ll long long

ll n, m, s;
ll y, ans;
ll w[200010], v[200010];
ll l[200010], r[200010];
ll qzh1[200010], qzh2[200010];

bool check(ll wq) {
	y = 0;
	memset(qzh1, 0, sizeof(qzh1));  
	memset(qzh2, 0, sizeof(qzh2));
	// 多测不清空，爆零两行泪 
	for (int i = 1; i <= n; i++) {
		if (w[i] > wq)  // > 过了检测通过线 
			qzh1[i] = qzh1[i - 1] + 1, qzh2[i] = qzh2[i - 1] + v[i]; // 前缀和加上 
		else
			qzh1[i] = qzh1[i - 1], qzh2[i] = qzh2[i - 1];  // 承继原来的前缀 
	}
	for (int i = 1; i <= m; i++) {
		int rrrr = r[i];
		int llll = l[i];
		y += (qzh1[rrrr] - qzh1[llll - 1]) * (qzh2[rrrr] - qzh2[llll - 1]);  // 直接照着式子，简单分析，就会发现这跟式子几乎一模一样/doge 
	}
	if (y > s)
		return 1;  // 判断差值大还是小，为了二分的更改做准备 
	else
		return 0;
}

int main() {
	cin >> n >> m >> s;
	for (int i = 1; i <= n; i++)
		cin >> w[i] >> v[i];
	for (int i = 1; i <= m; i++)
		cin >> l[i] >> r[i];
	int lll = 1;
	int rrr = 2000010;  
	ans = s;
	while (lll <= rrr) { // 二分答案 
		int mid = lll + (rrr - lll) / 2;  // 防爆ll， 原式子：(lll + rrr) / 2 
		if (check(mid))
			lll = mid + 1;
		else
			rrr = mid - 1;
			ans = min(ans, llabs(s - y)); // 为了防止爆int， 所以写llabs 
	}
	cout << ans;
}

#include<iostream>
#include<algorithm>
#include<cstdio>
#define int long long

using namespace std;

int n,m,s;
int w[200500],v[200500];
int l[200500],r[200500];

int sum_n[200500],sum_v[200500];

long long ans = 0;

void init()
{
    scanf("%lld%lld%lld",&n,&m,&s);
    for(int i = 1;i <= n; i++)
        scanf("%lld%lld",&w[i],&v[i]);
    for(int i = 1;i <= m; i++)
        scanf("%lld%lld",&l[i],&r[i]);
    
    return ;
}

long long check(int W)
{
    long long ans = 0;
    for(int i = 1;i <= n; i++)
    {
        if( W > w[i] )// 要用前缀和，不然会炸掉！！！
        {
            sum_n[i] = sum_n[i-1];
            sum_v[i] = sum_v[i-1]; 
        }
        else
        {
            sum_n[i] = sum_n[i-1] + 1;
            sum_v[i] = sum_v[i-1] + v[i]; 
        }
    }

    for(int i = 1;i <= m; i++)
    {
        long long a,b;
        a = sum_v[r[i]] - sum_v[l[i]-1];
        b = sum_n[r[i]] - sum_n[l[i]-1];
        ans += a*b;
    }

    return ans;
}

long long _abs(long long a)
{
    if( a > 0 )
        return a;
    return -a;
}

signed main()
{
    init();

    int left = 0,right = 1000000,mid;

    while( left <= right )
    {
        mid  = (left + right)>>1;
        if( check(mid) > s )
            left = mid + 1;
        else    
            right = mid - 1;
    }
    ans = _abs(check(left) - s);
    
    if( _abs(check(right) - s) < ans )
        ans = _abs(check(right) - s);
    
    printf("%lld",ans);
    return 0;
}