//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define int long long
typedef pair<int, int> T;
const int oo = 1e18, oo2 = 1e9+7, K = 30;
const int mod = 998244353;
#define ii __int128_t
void solve(){
int n, A, B; cin >> n >> A >> B;
//po obrzydliwych obliczeniach dostajemy ze co AB to sie powtarza + dziala na samplach
//jesli s \neq t mod AB, to daja rozne pary, nie wprost daja
//wiec musi s \equiv t mod B, ale wtedy s \neq t mod A (bo gdyby zachodzilo przystawanie, to mozna wymnozyc(?))
//czy to ze AB niekoniecznie wzglednie pierwsze przeszkadza w czyms?
//wiec wtedy pierwsza wartosc daje inna reszte
set<T>s;//przedzialy mod AB
ii CC = (ii)A*(ii)B/(ii)__gcd(A, B+1);
CC = min(CC, (ii)2*oo);
int C = CC;
for (int i = 0; i<n; i++){
int l, r; cin >> l >> r;
int L = l%C;
int R = r%C;
if (r-l+1 >= C) {
s.insert({0, C-1});
} else {
if (L <= R) s.insert({L, R});
else {
s.insert({L, C-1});
s.insert({0, R});
}
}
}
debug(s);
map<int, int>sweep;
for (auto [a, b]: s){
sweep[a]++;
sweep[b+1]--;
}
int prev = -1;
int open = 0, ans = 0;
for (auto [a, val]: sweep){
debug(a, val, open);
if (open) {
ans += a-prev;
}
open += val;
prev = a;
}
cout << ans << "\n";
}
int32_t main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
//cin >> t;
while (t--) solve();
return 0;
}
