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 mod = 1e9 + 7;
void print(int *A, int N) {
  for (int i = 0; i < N; i++) cerr << A[i] << " ";
  cerr << endl;
}
int solve(string s, int l, int r) {
  bool DONE[26];
  memset(DONE, 0, sizeof(DONE));
  int ANS = 0;
  for (int i = l; i <= r; i++) {
    if (DONE[s[i] - 97] == false) ANS++;
    DONE[s[i] - 97] = true;
  }
  return ANS;
}
int main() {
  int A, B, L, R;
  cin >> A >> B >> L >> R;
  if ((R - L) > 2 * A + B) {
    if (B >= A)
      cout << A + 1 << endl;
    else
      cout << 2 * A - B << endl;
  } else {
    int sum = A + B;
    int start = L % sum;
    if (start == 0) start = sum;
    int l = start;
    int r = R - L + start;
    string second = "";
    second += ".";
    for (int i = 1; i <= A; i++) {
      char ch = (i + 96);
      second += ch;
    }
    string T = second;
    for (int i = 0; i < B; i++) T += (char)(A + 96);
    string suf = T.substr(T.size() - A, A);
    int done[26];
    memset(done, 0, sizeof(done));
    for (int i = 0; i < suf.size(); i++) done[suf[i] - 97] = 1;
    int next = 0;
    for (int i = 0; i < A; i++) {
      while (done[next] == 1) next++;
      T += (char)(next + 97);
      next++;
    }
    int tmp;
    int ans = min(1000000, solve(T, l, r));
    cerr << T << endl;
    cerr << start << endl;
    for (int i = 0; i < min(B, max(0, A - start + 1)); i++) {
      second += (char)(i + start + 96);
    }
    for (int i = max(0, A - start + 1); i < B; i++) second += (char)(A + 96);
    cerr << second << endl;
    suf = second.substr(second.size() - A, A);
    memset(done, 0, sizeof(done));
    for (int i = 0; i < suf.size(); i++) done[suf[i] - 97] = 1;
    next = 0;
    for (int i = 0; i < A; i++) {
      while (done[next] == 1) next++;
      second += (char)(next + 97);
      next++;
    }
    for (int i = 0; i < min(B, max(0, A - start + 1)); i++) {
      second += (char)(i + start + 96);
    }
    for (int i = max(0, A - start + 1); i < B; i++) second += (char)(A + 96);
    cerr << second << endl;
    suf = second.substr(second.size() - A, A);
    memset(done, 0, sizeof(done));
    for (int i = 0; i < suf.size(); i++) done[suf[i] - 97] = 1;
    next = 0;
    for (int i = 0; i < A; i++) {
      while (done[next] == 1) next++;
      second += (char)(next + 97);
      next++;
    }
    for (int i = 0; i < min(B, max(0, A - start + 1)); i++) {
      second += (char)(i + start + 96);
    }
    for (int i = max(0, A - start + 1); i < B; i++) second += (char)(A + 96);
    cerr << second << endl;
    suf = second.substr(second.size() - A, A);
    memset(done, 0, sizeof(done));
    for (int i = 0; i < suf.size(); i++) done[suf[i] - 97] = 1;
    next = 0;
    for (int i = 0; i < A; i++) {
      while (done[next] == 1) next++;
      second += (char)(next + 97);
      next++;
    }
    for (int i = 0; i < min(B, max(0, A - start + 1)); i++) {
      second += (char)(i + start + 96);
    }
    for (int i = max(0, A - start + 1); i < B; i++) second += (char)(A + 96);
    cerr << second << endl;
    suf = second.substr(second.size() - A, A);
    memset(done, 0, sizeof(done));
    for (int i = 0; i < suf.size(); i++) done[suf[i] - 97] = 1;
    next = 0;
    for (int i = 0; i < A; i++) {
      while (done[next] == 1) next++;
      second += (char)(next + 97);
      next++;
    }
    for (int i = 0; i < min(B, max(0, A - start + 1)); i++) {
      second += (char)(i + start + 96);
    }
    for (int i = max(0, A - start + 1); i < B; i++) second += (char)(A + 96);
    cerr << second << endl;
    suf = second.substr(second.size() - A, A);
    memset(done, 0, sizeof(done));
    for (int i = 0; i < suf.size(); i++) done[suf[i] - 97] = 1;
    next = 0;
    for (int i = 0; i < A; i++) {
      while (done[next] == 1) next++;
      second += (char)(next + 97);
      next++;
    }
    cerr << second << endl;
    cerr << l << " " << r << endl;
    ans = min(ans, solve(second, l, r));
    cout << ans << endl;
  }
  return 0;
}
