C. Count Triangles

Like any unknown mathematician, Yuri has favourite numbers: AA, BB, CC, and DD, where A≤B≤C≤DA≤B≤C≤D. Yuri also likes triangles and once he thought: how many non-degenerate triangles with integer sides xx, yy, and zz exist, such that A≤x≤B≤y≤C≤z≤DA≤x≤B≤y≤C≤z≤D holds?

Yuri is preparing problems for a new contest now, so he is very busy. That's why he asked you to calculate the number of triangles with described property.

The triangle is called non-degenerate if and only if its vertices are not collinear.

Input

The first line contains four integers: AA, BB, CC and DD (1≤A≤B≤C≤D≤5⋅1051≤A≤B≤C≤D≤5⋅105) — Yuri's favourite numbers.

Output

Print the number of non-degenerate triangles with integer sides xx, yy, and zz such that the inequality A≤x≤B≤y≤C≤z≤DA≤x≤B≤y≤C≤z≤D holds.

Examples

input

Copy

1 2 3 4

output

Copy

4

input

Copy

1 2 2 5

output

Copy

3

input

Copy

500000 500000 500000 500000

output

Copy

1

Note

In the first example Yuri can make up triangles with sides (1,3,3)(1,3,3), (2,2,3)(2,2,3), (2,3,3)(2,3,3) and (2,3,4)(2,3,4).

In the second example Yuri can make up triangles with sides (1,2,2)(1,2,2), (2,2,2)(2,2,2) and (2,2,3)(2,2,3).

In the third example Yuri can make up only one equilateral triangle with sides equal to 5⋅1055⋅105.

 

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 5e5 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}
ll sum[N],cnt[N],tot;
int main()
{
    ll a, b, c, d;
    cin >> a >> b >> c >> d;
    ll l = -1, r = d - b+1,mx=min(d-c+1,c-b+1);
    while (r > l)
    {
        tot++;
        r--, l++;
        cnt[r] = cnt[l] = min(tot, mx);

    }
    sum[0] = cnt[0];
    rep(i, 1, d)
    {
        sum[i] = sum[i - 1] + cnt[i];
    }
    ll ans = 0;
    rep(i,a,b)
    {
        ans += sum[i - 1];
    }
    cout << ans << endl;
    return 0;
}