Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.

All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.

Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").

The players take turns appending letters to string s. Mister B moves first.

Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.

Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.

Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.

Input

First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.

Output

Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.

Examples

Input

1 1 1 8


Output

2

Input

4 2 2 6


Output

3

Input

3 7 4 6


Output

1

Note

In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.

In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.

In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
int num[10][100];
int used[100];
int main() {
  int n, m;
  int i, j, k;
  int a, b, c;
  cin >> n >> m >> a >> b;
  if (b - a > 3 * (n + m)) {
    c = n + max(n - m, 1);
    cout << c << endl;
    return 0;
  }
  for (i = 0; i < n + m; i++) {
    if (i < n)
      num[0][i] = i;
    else
      num[0][i] = n - 1;
  }
  for (i = 1; i < 5; i++) {
    memset(used, false, sizeof(used));
    for (j = m; j < n + m; j++) {
      used[num[i - 1][j]] = true;
    }
    k = 0;
    for (j = 0; j < n; j++) {
      while (used[k]) k++;
      num[i][j] = k;
      used[k] = true;
    }
    for (j = 0; j < m; j++) {
      num[i][n + j] = num[i][n - 1];
    }
  }
  a--, b--;
  c = a / (n + m) * (n + m);
  a -= c, b -= c;
  if (a >= n && n > m) {
    for (i = 0; i < m; i++) {
      num[0][n + i] = num[0][m];
    }
  }
  memset(used, false, sizeof(used));
  int s = 0;
  for (i = a; i <= b; i++) {
    c = num[i / (n + m)][i % (n + m)];
    if (used[c]) continue;
    used[c] = true;
    s++;
  }
  cout << s << endl;
}
