/*input
3 1
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll sub(ll a,ll b){
ll x=a-b;
while(x<0) x+=MOD;
while(x>MOD) x-=MOD;
return x;
}
ll mult(ll a,ll b){
return a*b%MOD;
}
ll mypow(ll a,ll b){
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1) res=(res*a)%MOD;
a=(a*a)%MOD;
b>>=1;
}
return res;
}
typedef pair<ll,ll> ii;
int n;
ll a, b, t;
vector<ii> p;
const ll MAXT = 1e18+18;
int main(){
ios::sync_with_stdio(false),cin.tie(0);
cin>>n>>a>>b;
p.resize(n);
for(int i=0;i<n;++i){
cin>>p[i].f>>p[i].s;
}
a = a/__gcd(a, b+1);
if(b>MAXT/a){ //every t has a distinct pair
ll sol = 0;
for(auto i:p){
sol += i.second-i.first+1;
}
cout<<sol;
}else{ //pairs repeat every t moments
ll t = a*b;
vector<ii> p2;
for(auto i:p){
if(i.first/t==i.second/t){
p2.push_back(ii(i.first%t, i.second%t));
}else if(i.first/t+1==i.second/t){
p2.push_back(ii(i.first%t, t-1));
p2.push_back(ii(0, i.second%t));
}else{
cout<<t;
return 0;
}
}
sort(p2.begin(), p2.end());
ll s, e, sol=0;
for(int i=0, j=0;i<p2.size();i=j){
s=p2[i].first;e=p2[i].second;
for(;j<p2.size() && p2[j].first<=e;++j){
e=max(e, p2[j].second);
}
sol+=e-s+1;
}
cout<<sol;
}
}
