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 long long UNDEF = -1;
const long long INF = 1e18;
template <typename T>
inline bool chkmax(T& aa, T bb) {
  return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T& aa, T bb) {
  return aa > bb ? aa = bb, true : false;
}
static char stdinBuffer[1024];
static char* stdinDataEnd = stdinBuffer + sizeof(stdinBuffer);
static const char* stdinPos = stdinDataEnd;
void readAhead(size_t amount) {
  size_t remaining = stdinDataEnd - stdinPos;
  if (remaining < amount) {
    memmove(stdinBuffer, stdinPos, remaining);
    size_t sz = fread(stdinBuffer + remaining, 1,
                      sizeof(stdinBuffer) - remaining, stdin);
    stdinPos = stdinBuffer;
    stdinDataEnd = stdinBuffer + remaining + sz;
    if (stdinDataEnd != stdinBuffer + sizeof(stdinBuffer)) *stdinDataEnd = 0;
  }
}
int readInt() {
  readAhead(16);
  int x = 0;
  bool neg = false;
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  if (*stdinPos == '-') {
    ++stdinPos;
    neg = true;
  }
  while (*stdinPos >= '0' && *stdinPos <= '9') {
    x *= 10;
    x += *stdinPos - '0';
    ++stdinPos;
  }
  return neg ? -x : x;
}
char readCh() {
  readAhead(16);
  while (*stdinPos == ' ' || *stdinPos == '\n') ++stdinPos;
  char ans = *stdinPos;
  ++stdinPos;
  return ans;
}
const int mn = 1e5 + 4;
bool s[26];
int t[mn];
int DEL = 0;
int g(int a, int b, int l, int r) {
  for (int i = 0; i < a; i++) {
    t[i] = i;
  }
  int x = a;
  while (x <= r) {
    int use = t[x - 1];
    for (int i = 0; i < a - b; i++) {
      chkmin(use, t[x - 1 - i]);
    }
    for (int i = 0; i < b; i++) {
      t[x] = use;
      x++;
    }
    memset(s, 0, sizeof s);
    for (int i = 0; i < a; i++) {
      s[t[x - 1 - i]] = 1;
    }
    int c = 0;
    for (int i = 0; i < a; i++) {
      while (s[c]) c++;
      t[x] = c;
      x++;
      c++;
    }
  }
  memset(s, 0, sizeof s);
  for (int x = l; x <= r; x++) s[t[x]] = 1;
  int ans = 0;
  for (int c = 0; c < 26; c++) ans += s[c];
  return ans;
}
int f(int a, int b, int l, int r, int use, int b4) {
  for (int i = 0; i < a; i++) {
    t[i] = i;
  }
  int x = a;
  while (x <= r) {
    for (int i = 0; i < b; i++) {
      if (x < l)
        t[x] = b4;
      else
        t[x] = use;
      x++;
    }
    memset(s, 0, sizeof s);
    for (int i = 0; i < a; i++) {
      s[t[x - 1 - i]] = 1;
    }
    int c = 0;
    for (int i = 0; i < a; i++) {
      while (s[c]) c++;
      t[x] = c;
      x++;
      c++;
    }
  }
  memset(s, 0, sizeof s);
  for (int x = l; x <= r; x++) s[t[x]] = 1;
  int ans = 0;
  for (int c = 0; c < 26; c++) ans += s[c];
  return ans;
}
const int D = 300;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int a, b, l, r;
  scanf("%d%d%d%d", &a, &b, &l, &r);
  --l;
  --r;
  int dl = l / (a + b);
  if (dl >= D) {
    l -= (dl - D) * (a + b);
    r -= (dl - D) * (a + b);
  }
  int dx = (r - l) / (a + b);
  DEL = 0;
  if (dx >= D) {
    r -= (dx - D) * (a + b);
    DEL = dx - D;
  }
  int ans = g(a, b, l, r);
  for (int x = 0; x < 26; x++)
    for (int y = 0; y < 26; y++) chkmin(ans, f(a, b, l, r, x, y));
  printf("%d\n", ans);
}
