// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define sz(x) int(x.size())
using ll = long long;
template<class T> using V = vector<T>;
using vl = V<ll>;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N; ll A, B; cin >> N >> A >> B;
ll K = A / __gcd(B + 1, A);
if (K > ll(1e18) / B) K = -1;
else K *= B;
V<array<ll, 2>> E;
auto add = [&](ll l, ll r) {
E.pb(array<ll, 2>{l, +1});
E.pb(array<ll, 2>{r + 1, -1});
};
ll ans = 0;
for(int i = 0; i < N; i++) {
ll l, r; cin >> l >> r;
ll len = r - l + 1;
if (K == -1) ans += len;
if (len >= K) add(0, K - 1);
else {
l %= K, r %= K;
if (l <= r) add(l, r);
else add(l, K - 1), add(0, r);
}
}
if (K == -1) {
cout << ans << nl;
exit(0-0);
}
int have = 0;
sort(begin(E), end(E));
for(int i = 0; i < sz(E); i++) {
int j = i; while(j < sz(E) && E[j][0] == E[i][0]) {
have += E[j][1]; j++;
}
if (have && j < sz(E)) {
ll amt = E[j][0] - E[i][0];
// cout << E[j][0] << " " << E[i][0] << endl;
ans += amt;
}
i = j - 1;
}
cout << ans << nl;
exit(0-0);
}
