#include <bits/stdc++.h>
using namespace std;
class strangeDevice{
private:
using ll=long long;
using i128=__int128;
public:
long long count(long long A, long long B, vector<long long> left, vector<long long> right) {
long long ans{0ll};
i128 F = i128(A / __gcd(A, B + 1)) * B;
vector<pair<ll,ll> > rgs, rgx;
for(int i{0}; i < (int)left.size(); ++i){
long long l{left[i]}, r{right[i]};
long long len{r - l + 1};
if(len>=F) return F;
l %= F, r %= F;
if(l <= r) rgs.emplace_back(l, r);
else rgs.emplace_back(l, F - 1),rgs.emplace_back(0, r);
}
sort(rgs.begin(), rgs.end());
for(auto [l,r]: rgs){
if(rgx.empty() || rgx.back().second < l ) rgx.emplace_back(l, r);
else rgx.back().second = max(rgx.back().second, r);
}
for(auto [l,r]: rgx) ans += r - l + 1;
return ans;
}
}solution;
int main() {
cin.tie(0) -> sync_with_stdio(0);
int lengthOfL;
long long A, B;
cin >> lengthOfL >> A >> B;
vector<long long> L(lengthOfL),R(lengthOfL);
for(int i = 0; i < lengthOfL; i++ ){
cin >> L[i] >> R[i];
}
cout << solution.count(A, B, L, R) << '\n';
return 0;
}
/*
这类题目可以考虑找循环节
t1=xB+y,t2=x1B+y
x1B+y+[(x1B+y)/B]=x2B+y+[(x2B+y)/B]
x1B+y+x1=x2B+y+x2
(x1-x2)(B+1)=0 (mod A)
x1-x2=K* A/gcd(A,B+1)
t1-t2=K* AB/gcd(A,B+1)
∴循环节(最小正周期) F=AB/gcd(A,B+1)
转化成：有一些区间，求区间的并 %F 意义下有多少个不同的数
len>=F: ans=F
否则转化成了区间覆盖操作
*/
