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 double EPS = (double)1e-9;
const double PI = (double)acos(-1.0);
int irand(int lo, int hi) {
  return (((double)rand()) / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
string toString(long long x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
long long toNumber(string S) {
  long long ret;
  sscanf(S.c_str(), "%lld", &ret);
  return ret;
}
const int INF = (int)2e9;
const long long MOD = (long long)1e9 + 7;
int main() {
  int a, b, l, r;
  scanf("%d %d %d %d", &a, &b, &l, &r);
  int jeda = (a - b > 0 ? a - b : 1);
  int range = (r - l + 1);
  if (range > 2 * (a + b)) l = 1, r = 2 * (a + b);
  int awal = (l - 1) % (2 * (a + b));
  int akir = (r - 1) % (2 * (a + b));
  if (awal >= akir && l < r) akir += 2 * (a + b);
  assert(awal <= akir);
  int s[205];
  for (int i = 0; i < 203; i++) s[i] = -1;
  for (int i = 0; i <= a - 1; i += 1) s[i] = s[i + a + b] = i;
  for (int i = 0; i <= b - 1; i += 1) s[a + i] = s[a - 1];
  for (int i = a + b + a - jeda; i <= a + b + a - 1; i += 1) s[i] += jeda;
  for (int i = 0; i <= b - 1; i += 1) s[a + b + a + i] = s[a + b + a - 1];
  for (int i = 2 * (a + b); i <= 8 * (a + b); i += 1) s[i] = s[i - 2 * (a + b)];
  set<int> temp;
  temp.clear();
  for (int i = awal; i <= akir; i += 1) temp.insert(s[i]);
  for (int i = awal; i <= akir; i += 1) assert(s[i] >= 0);
  int res = (int)temp.size();
  if (jeda > 1 && a == awal && akir == a + b + a + b + a - 2) res--;
  fprintf(stderr, "     << %d %d\n", awal, akir);
  fprintf(stderr, "     << ");
  for (int i = 0; i < (a + b) << 2; i++) fprintf(stderr, "%d ", s[i]);
  fprintf(stderr, "\n");
  printf("%d\n", res);
  return 0;
}
