#include <bits/stdc++.h>
using namespace std;
int a, b, l, r;
char s[1005];
bool v[30];
int pp;
void get(char op) {
  for (int i = a + 1; i <= (a + b); i++) s[i] = op;
  memset(v, 0, sizeof(v));
  for (int i = a + b, j = 0; j < a; j++, i--) v[s[i] - 'a'] = 1;
  for (int i = a + b + 1; i <= (a + a + b); i++) {
    for (int j = 0; j < 26; j++) {
      if (!v[j]) {
        v[j] = 1;
        s[i] = 'a' + j;
        break;
      }
    }
  }
  for (int i = (a + a + b + 1); i <= (a + a + b + b); i++) s[i] = s[a + a + b];
  for (int i = (a + a + b + b + 1); i <= 4 * (a + b); i++)
    s[i] = s[i - a - a - b - b];
}
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  for (int i = 1; i <= a; i++) s[i] = 'a' + i - 1;
  if (r - l + 1 >= 2 * (a + b)) {
    if (a <= b)
      printf("%d\n", a + 1);
    else
      printf("%d\n", 2 * a - b);
  } else {
    pp = a + a + b + b;
    l = l % pp;
    if (!l) l = pp;
    r = r % pp;
    if (!r) r = pp;
    set<char> st;
    int ans = 1e9;
    int temp;
    for (int i = 0; i < a; i++) {
      get('a' + i);
      st.clear();
      if (l > r) {
        for (int i = l; i <= 2 * (a + b); i++) st.insert(s[i]);
        for (int i = 1; i <= r; i++) st.insert(s[i]);
        temp = st.size();
      } else {
        for (int i = l; i <= r; i++) st.insert(s[i]);
        temp = st.size();
      }
      ans = min(ans, temp);
    }
    printf("%d\n", ans);
  }
}

### Prompt

In Cpp, your task is to solve 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;
char s[1005];
bool v[30];
int pp;
void get(char op) {
  for (int i = a + 1; i <= (a + b); i++) s[i] = op;
  memset(v, 0, sizeof(v));
  for (int i = a + b, j = 0; j < a; j++, i--) v[s[i] - 'a'] = 1;
  for (int i = a + b + 1; i <= (a + a + b); i++) {
    for (int j = 0; j < 26; j++) {
      if (!v[j]) {
        v[j] = 1;
        s[i] = 'a' + j;
        break;
      }
    }
  }
  for (int i = (a + a + b + 1); i <= (a + a + b + b); i++) s[i] = s[a + a + b];
  for (int i = (a + a + b + b + 1); i <= 4 * (a + b); i++)
    s[i] = s[i - a - a - b - b];
}
int main() {
  scanf("%d%d%d%d", &a, &b, &l, &r);
  for (int i = 1; i <= a; i++) s[i] = 'a' + i - 1;
  if (r - l + 1 >= 2 * (a + b)) {
    if (a <= b)
      printf("%d\n", a + 1);
    else
      printf("%d\n", 2 * a - b);
  } else {
    pp = a + a + b + b;
    l = l % pp;
    if (!l) l = pp;
    r = r % pp;
    if (!r) r = pp;
    set<char> st;
    int ans = 1e9;
    int temp;
    for (int i = 0; i < a; i++) {
      get('a' + i);
      st.clear();
      if (l > r) {
        for (int i = l; i <= 2 * (a + b); i++) st.insert(s[i]);
        for (int i = 1; i <= r; i++) st.insert(s[i]);
        temp = st.size();
      } else {
        for (int i = l; i <= r; i++) st.insert(s[i]);
        temp = st.size();
      }
      ans = min(ans, temp);
    }
    printf("%d\n", ans);
  }
}

```