n, x = map(int, input().split())
now = 1
ans = 0
for i in range(n):
l, r = map(int, input().split())
ans += (l - now) % x + r - l
now = r + 1
print(ans + n)
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,x;
cin>>n>>x;
int last=1,ans=0;
for (int i=0;i<n;i++)
{
int l,r;//We Take the first minutes and last minutes as input.
cin>>l>>r;
ans+=(r-l+1)+(l-last)%x;//First We calculate Total duration then We calculate the
//The amount of time left after Skipping.like if the difference b/w two best
//Moments is 3 and skipping allowed is 2 then 1 minutes will be remaining which we
//have to watch.
last=r+1;
}
cout<<ans<<endl;
return 0;
}
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490A - Team Olympiad | 233A - Perfect Permutation |
1360A - Minimal Square | 467A - George and Accommodation |
893C - Rumor | 227B - Effective Approach |
1534B - Histogram Ugliness | 1611B - Team Composition Programmers and Mathematicians |
110A - Nearly Lucky Number | 1220B - Multiplication Table |
1644A - Doors and Keys | 1644B - Anti-Fibonacci Permutation |
1610A - Anti Light's Cell Guessing | 349B - Color the Fence |
144A - Arrival of the General | 1106A - Lunar New Year and Cross Counting |
58A - Chat room | 230A - Dragons |
200B - Drinks | 13A - Numbers |
129A - Cookies | 1367B - Even Array |