/*
#include <stdio.h>
#include <math.h>
int n, p;
int x[10001]={0}, y[10001]={0}, pi[10001]={0};
double r[10001]={0};
double arr[10001]={0};
int chk = 0;
void calc(double left, double right)
{
if (left > right)
{
return;
}
double mid = (left+right)/2;
int sum;
for (int i=0; i<n; i++)
{
sum = 0;
if (r[i] < mid)
{
sum += pi[i];
}
}
if (p+sum > 1000000)
{
printf("1: %lf\n", mid);
arr[chk++] = mid;
calc(left, mid-1);
}
else if (p+sum == 1000000)
{
printf("2: %lf\n", mid);
arr[chk++] = mid;
return;
}
else if (p+sum < 1000000)
{
printf("3: %lf\n", mid);
calc(mid+1, right);
}
}
int main()
{
double temp;
scanf("%d %d", &n, &p);
for (int i=0; i<n; i++)
{
scanf("%d %d %d", &x[i], &y[i], &pi[i]);
r[i] = sqrt(x[i]*x[i] + y[i]*y[i]);
}
int sum = 0;
for (int i=0; i<n; i++)
{
sum += pi[i];
}
if (sum+p < 1000000)
{
printf("-1");
return 0;
}
calc(0, 10000*sqrt(2));
for (int i=0; i<chk-1; i++)
{
printf("%.3lf\n", arr[i]);
}
for (int i=0; i<chk-2; i++)
{
for (int j=0; j<chk-i-2; j++)
{
if (arr[j] > arr[j+1])
{
temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
}
printf("%.3lf", arr[0]);
}
*/
/*
#include <stdio.h>
int memo[20000] = {0};
int count(int n)
{
if (n==1)
{
return memo[n] = 1;
}
if (n==2)
{
return memo[n] = 2;
}
if (memo[n] != 0)
{
return memo[n];
}
else
{
return memo[n] = count(n-1)%100000007 + count(n-2)%100000007;
}
}
int main()
{
int n;
scanf("%d", &n);
int result = count(n)%100000007;
printf("%d", result);
}
*/
/*
#include <stdio.h>
int memo[20000] = {0};
int memo2[20000] = {0};
int count(int n)
{
if (n==1)
{
return memo[n] = 1;
}
else if (n==2)
{
return memo[n] = 2;
}
else if (n==3)
{
return memo[n]= 3;
}
if (memo[n] != 0)
{
return memo[n];
}
else
{
memo[n] = count(n-1) % 100000007 + count(n-1) %100000007;
}
}
int count2(int n)
{
if (n==1)
{
return memo2[n] = 0;
}
else if (n==2)
{
return memo2[n] = 1;
}
else if (n==3)
{
return memo2[n] = 2;
}
if (memo2[n] != 0)
{
return memo2[n];
}
else
{
memo2[n] = (count2(n-1) * 2 + 1) % 100000007;
}
}
int main()
{
int n;
scanf("%d", &n);
int result = count(n) % 100000007 + count2(n) % 100000007;
printf("%d", result);
}
*/
/*
#include <stdio.h>
int memo[200000] = {0};
int count(int n)
{
if (n==1)
return memo[n] = 1;
else if (n==2)
return memo[n] = 3;
if (memo[n] != 0)
return memo[n];
else
return memo[n] = count(n-1)%100007 + (count(n-2)%100007) * 2;
}
int main()
{
int n;
scanf("%d", &n);
int result = count(n)%100007;
printf("%d", result);
}
*/
#include <stdio.h>
int d[2001] = {0};
void count(int start, int end)
{
}
int main()
{
int d1, d2;
int sum = 0;
scanf("%d %d", &d1, &d2);
for (int i=1; i<=2000; i++)
{
d[i] = i;
}
count(d1, d2);
for (int i=d1; i<=d2; i++)
{
sum += d[i];
}
}