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#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 100010;
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struct Node {
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int l, r;
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const bool operator<(const Node &b) const { // 按每个区间的左端点从小到大排序
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return l < b.l;
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}
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} range[N];
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int n; // n个区间
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int st, ed; // 开始端点,结束端点
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int res; // 选择的区间数
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int main() {
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// 输入
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cin >> st >> ed >> n;
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for (int i = 0; i < n; i++) {
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int l, r;
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cin >> l >> r;
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range[i] = {l, r};
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}
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// 1、按左端点从小到大排序
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sort(range, range + n);
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// 2、遍历每个区间,注意这里的i没有++,因为可能一次跳过多个区间
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for (int i = 0; i < n;) {
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int j = i;
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int r = -INF; // 预求最大,先设最小
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// 3、双指针,从当前区间开始向后,找出覆盖start起点的区间,就是让区间尽可能的长
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while (j < n && range[j].l <= st) {
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r = max(r, range[j].r); // 找出右端最长的那个区间
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j++;
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}
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// 4、如果没有找到,表示出现了空隙
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if (r < st) {
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cout << -1 << endl;
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exit(0);
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}
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// 5、如果找到,多找出了一个区间
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res++;
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// 6、如果已经完整覆盖,输出
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if (r >= ed) {
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cout << res << endl;
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exit(0);
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}
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// 7、更新迭代起点
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st = r;
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// 指针跳跃
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i = j;
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}
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// 7、如果运行到这里,表示无法覆盖掉所有点
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cout << -1 << endl;
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return 0;
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}
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