https://www.luogu.com.cn/problem/P3963

 #define mid ((l+r)>>1)
const int N = 200010;
struct node
{
    int cj, jj;
}a[N];
int b[N];
int cmp(node a, node b)
{
    return a.cj < b.cj;
}
int ls[N * 40], rs[N * 40], sum[N * 40], ans[N * 40], root[N], idx;
void build(int& u, int l, int r)
{
    u = ++idx;
    if (l == r) return;
    build(ls[u], l, mid);
    build(rs[u], mid + 1, r);
}
void push_up(int u)
{
    sum[u] = sum[ls[u]] + sum[rs[u]];
    ans[u] = ans[ls[u]] + ans[rs[u]];
}
void add(int& u, int v, int l, int r, int p, int k)
{
    u = ++idx;
    ls[u] = ls[v];
    rs[u] = rs[v];
    sum[u] = sum[v];
    ans[u] = ans[v];
    if (l == r)
    {
        sum[u]++;
        ans[u] += k;
        return;
    }
    if (p <= mid) add(ls[u], ls[v], l, mid, p, k);
    else add(rs[u], rs[v], mid + 1, r, p, k);
    push_up(u);
}
ll find(int u, int v, int l, int r, int k)
{
    ll res = 0;
    if (l == r) return k * b[l];    //叶子节点l存的是这个数离散化后的值，但是到了叶子节点的时候k值不一定为1，因为l的sum（数字个数）值不一定为1
    int s = sum[ls[u]] - sum[ls[v]];         //所以要乘k，b[l]是原数组的值
    if (k <= s) res += find(ls[u], ls[v], l, mid, k);
    else
    {
        res += ans[ls[u]] - ans[ls[v]];
        res += find(rs[u], rs[v], mid + 1, r, k - s);
    }
    return res;
}
void solve()
{
    int n, c, f;
    cin >> n >> c >> f;
    for (int i = 1; i <= c; i++) cin >> a[i].cj >> a[i].jj, b[i] = a[i].jj;
    sort(a + 1, a + 1 + c, cmp);
    sort(b + 1, b + 1 + c);
    int len = unique(b + 1, b + 1 + c) - b - 1;
    build(root[0], 1, len);
    for (int i = 1; i <= c; i++)
    {
        int x = lower_bound(b + 1, b + 1 + len, a[i].jj) - b;
        add(root[i], root[i-1], 1, len, x, a[i].jj);
    }
    int k = n / 2;
    int ok = 0;
    for (int i = c - k; i >= 1 + k; i--)
    {
        ll cnty = find(root[c], root[i], 1, len, k);
        ll cntz = find(root[i - 1], root[0], 1, len, k);
        if (cntz + cnty + a[i].jj <= f)
        {
            cout << a[i].cj << '\n';
            ok = 1;
            break;
        }
    }
    if (ok == 0) cout << "-1" << '\n';
}
int main()
{
    ios::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}