#include <iostream>
#include <cstring>
#include <algorithm>
#include <deque>

using namespace std;

typedef long long LL;

const int N = 500010;

int n, d, k;
int x[N], w[N];
LL f[N];

bool check(int mid)
{
    int l = max(1, d - mid), r = d + mid;
    LL res = 0;
    deque<int> dq; // 使用 deque 代替手动维护的单调队列
    memset(f, -0x3f, sizeof f);
    f[0] = 0;

    for (int i = 1, j = 0, k = 0; i <= n; i++)
    {
        // 更新队列：确保窗口的左边界 >= l
        while (x[i] - x[k] >= l)
        {
            while (!dq.empty() && f[dq.back()] <= f[k])
                dq.pop_back();
            dq.push_back(k++);
        }

        // 移动 j，确保窗口的右边界 <= r
        while (x[i] - x[j] > r) j++;

        // 移除过期的元素
        while (!dq.empty() && dq.front() < j)
            dq.pop_front();

        // 如果队列非空，则更新 f[i]
        if (!dq.empty())
            f[i] = f[dq.front()] + w[i];

        // 更新结果
        res = max(res, f[i]);
    }

    return res >= k;
}

int main()
{
    scanf("%d%d%d", &n, &d, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &x[i], &w[i]);

    int l = 0, r = 1e9;
    while (l < r)
    {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }

    // 边界检查
    if (!check(r)) r = -1;
    printf("%d\n", r);

    return 0;
}