#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
int flag[30];
int s[200];
int tol;
void init1() {
  memset(s, 0, sizeof(s));
  int i, j, tmp;
  int mp[30];
  for (i = 1; i <= a; i++) s[i] = i - 1;
  s[a + 1] = s[b + 1];
  for (i = a + 2; i <= a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = a + b + 1; i <= a + b + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[a + b + a + 1] = s[a + b + b + 1];
  for (i = a + b + a + 2; i <= a + b + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 2 + 1; i <= (a + b) * 2 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 2 + a + 1] = s[(a + b) * 2 + b + 1];
  for (i = (a + b) * 2 + a + 2; i <= (a + b) * 2 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 3 + 1; i <= (a + b) * 3 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 3 + a + 1] = s[(a + b) * 3 + b + 1];
  for (i = (a + b) * 3 + a + 2; i <= (a + b) * 3 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  tol = 0;
  for (i = 1; i <= (a + b) * 2; i++) {
    if (mp[s[i]] == 0) tol++;
    mp[s[i]] = 1;
  }
}
void init2() {
  int i, j, tmp;
  int mp[30];
  for (i = 1; i <= a; i++) s[i] = i - 1;
  s[a + 1] = s[a];
  for (i = a + 2; i <= a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = a + b + 1; i <= a + b + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[a + b + a + 1] = s[a + b + a];
  for (i = a + b + a + 2; i <= a + b + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 2 + 1; i <= (a + b) * 2 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 2 + a + 1] = s[(a + b) * 2 + a];
  for (i = (a + b) * 2 + a + 2; i <= (a + b) * 2 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 3 + 1; i <= (a + b) * 3 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 3 + a + 1] = s[(a + b) * 3 + a];
  for (i = (a + b) * 3 + a + 2; i <= (a + b) * 3 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  tol = 0;
  for (i = 1; i <= (a + b) * 2; i++) {
    if (mp[s[i]] == 0) tol++;
    mp[s[i]] = 1;
  }
}
int main() {
  int i, j, tmp;
  cin >> a >> b >> l >> r;
  init1();
  if (r - l + 1 >= (a + b) * 2) {
    cout << tol << endl;
    return 0;
  }
  l = (l - 1) % (2 * (a + b)) + 1;
  r = (r - 1) % (2 * (a + b)) + 1;
  if (r < l) r += 2 * (a + b);
  int ans1 = 0, ans2 = 0;
  init1();
  memset(flag, 0, sizeof(flag));
  for (i = l; i <= r; i++) {
    tmp = i;
    flag[s[tmp]] = 1;
  }
  for (i = 0; i < 26; i++)
    if (flag[i]) ans1++;
  init2();
  memset(flag, 0, sizeof(flag));
  for (i = l; i <= r; i++) {
    tmp = i;
    flag[s[tmp]] = 1;
  }
  for (i = 0; i < 26; i++)
    if (flag[i]) ans2++;
  cout << min(ans1, ans2) << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
int flag[30];
int s[200];
int tol;
void init1() {
  memset(s, 0, sizeof(s));
  int i, j, tmp;
  int mp[30];
  for (i = 1; i <= a; i++) s[i] = i - 1;
  s[a + 1] = s[b + 1];
  for (i = a + 2; i <= a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = a + b + 1; i <= a + b + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[a + b + a + 1] = s[a + b + b + 1];
  for (i = a + b + a + 2; i <= a + b + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 2 + 1; i <= (a + b) * 2 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 2 + a + 1] = s[(a + b) * 2 + b + 1];
  for (i = (a + b) * 2 + a + 2; i <= (a + b) * 2 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 3 + 1; i <= (a + b) * 3 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 3 + a + 1] = s[(a + b) * 3 + b + 1];
  for (i = (a + b) * 3 + a + 2; i <= (a + b) * 3 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  tol = 0;
  for (i = 1; i <= (a + b) * 2; i++) {
    if (mp[s[i]] == 0) tol++;
    mp[s[i]] = 1;
  }
}
void init2() {
  int i, j, tmp;
  int mp[30];
  for (i = 1; i <= a; i++) s[i] = i - 1;
  s[a + 1] = s[a];
  for (i = a + 2; i <= a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = a + b + 1; i <= a + b + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[a + b + a + 1] = s[a + b + a];
  for (i = a + b + a + 2; i <= a + b + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 2 + 1; i <= (a + b) * 2 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 2 + a + 1] = s[(a + b) * 2 + a];
  for (i = (a + b) * 2 + a + 2; i <= (a + b) * 2 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  for (j = i - a; j <= i - 1; j++) mp[s[j]] = 1;
  for (i = (a + b) * 3 + 1; i <= (a + b) * 3 + a; i++) {
    for (j = 0; j < 26; j++)
      if (mp[j] == 0) {
        tmp = j;
        mp[tmp] = 1;
        break;
      }
    s[i] = tmp;
  }
  s[(a + b) * 3 + a + 1] = s[(a + b) * 3 + a];
  for (i = (a + b) * 3 + a + 2; i <= (a + b) * 3 + a + b; i++) s[i] = s[i - 1];
  memset(mp, 0, sizeof(mp));
  tol = 0;
  for (i = 1; i <= (a + b) * 2; i++) {
    if (mp[s[i]] == 0) tol++;
    mp[s[i]] = 1;
  }
}
int main() {
  int i, j, tmp;
  cin >> a >> b >> l >> r;
  init1();
  if (r - l + 1 >= (a + b) * 2) {
    cout << tol << endl;
    return 0;
  }
  l = (l - 1) % (2 * (a + b)) + 1;
  r = (r - 1) % (2 * (a + b)) + 1;
  if (r < l) r += 2 * (a + b);
  int ans1 = 0, ans2 = 0;
  init1();
  memset(flag, 0, sizeof(flag));
  for (i = l; i <= r; i++) {
    tmp = i;
    flag[s[tmp]] = 1;
  }
  for (i = 0; i < 26; i++)
    if (flag[i]) ans1++;
  init2();
  memset(flag, 0, sizeof(flag));
  for (i = l; i <= r; i++) {
    tmp = i;
    flag[s[tmp]] = 1;
  }
  for (i = 0; i < 26; i++)
    if (flag[i]) ans2++;
  cout << min(ans1, ans2) << endl;
  return 0;
}

```