There are N pinholes on the xy-plane. The i-th pinhole is located at (x_i,y_i).
We will denote the Manhattan distance between the i-th and j-th pinholes as d(i,j)(=|x_i-x_j|+|y_i-y_j|).
You have a peculiar pair of compasses, called Manhattan Compass.
This instrument always points at two of the pinholes.
The two legs of the compass are indistinguishable, thus we do not distinguish the following two states: the state where the compass points at the p-th and q-th pinholes, and the state where it points at the q-th and p-th pinholes.
When the compass points at the p-th and q-th pinholes and d(p,q)=d(p,r), one of the legs can be moved so that the compass will point at the p-th and r-th pinholes.
Initially, the compass points at the a-th and b-th pinholes.
Find the number of the pairs of pinholes that can be pointed by the compass.

-----Constraints-----
 - 2≦N≦10^5
 - 1≦x_i, y_i≦10^9
 - 1≦a < b≦N
 - When i ≠ j, (x_i, y_i) ≠ (x_j, y_j)
 - x_i and y_i are integers.

-----Input-----
The input is given from Standard Input in the following format:
N a b
x_1 y_1
:
x_N y_N

-----Output-----
Print the number of the pairs of pinholes that can be pointed by the compass.

-----Sample Input-----
5 1 2
1 1
4 3
6 1
5 5
4 8

-----Sample Output-----
4

Initially, the compass points at the first and second pinholes.
Since d(1,2) = d(1,3), the compass can be moved so that it will point at the first and third pinholes.
Since d(1,3) = d(3,4), the compass can also point at the third and fourth pinholes.
Since d(1,2) = d(2,5), the compass can also point at the second and fifth pinholes.
No other pairs of pinholes can be pointed by the compass, thus the answer is 4.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\ninput = sys.stdin.readline\nfrom collections import defaultdict\nimport bisect\n\nINF = 10 ** 12\n\nN,a,b = map(int,input().split())\nX_to_Y = defaultdict(lambda: [-INF,INF])\nY_to_X = defaultdict(lambda: [-INF,INF])\nfor i in range(1,N+1):\n    x,y = map(int,input().split())\n    x,y = x+y,x-y\n    X_to_Y[x].append(y)\n    Y_to_X[y].append(x)\n    if i == a:\n        sx,sy = x,y\n    elif i == b:\n        tx,ty = x,y\n\nR = max(abs(sx-tx), abs(sy-ty))\n\nfor key, arr in X_to_Y.items():\n    arr.sort()\n    L = len(arr)\n    move_right = list(range(1,L)) + [None]\n    X_to_Y[key] = [arr,move_right]\n    \nfor key, arr in Y_to_X.items():\n    arr.sort()\n    L = len(arr)\n    move_right = list(range(1,L)) + [None]\n    Y_to_X[key] = [arr,move_right]\n\nequiv_class = set([(sx,sy), (tx,ty)])\n\nanswer = 0\ntask = [(sx,sy), (tx,ty)]\n\nwhile task:\n    x,y = task.pop()\n    # 既出の元も含めて辺の個数を数える\n    # 未出の元を同値に追加\n    # x,y両方満たすものは、x側にだけ入れる\n    for x1 in [x-R,x+R]:\n        if not (x1 in X_to_Y):\n            continue\n        arr, move_right = X_to_Y[x1]\n        left = bisect.bisect_left(arr, y-R)\n        right = bisect.bisect_right(arr, y+R)\n        # [left, right) が辺で結べるターゲット\n        answer += right - left # 辺の個数\n        i = left\n        while i < right:\n            y1 = arr[i]\n            if (x1,y1) not in equiv_class:\n                equiv_class.add((x1,y1))\n                task.append((x1,y1))\n            # なるべく、既に居ない元は見ないで済ませるように\n            next_i = move_right[i]\n            if next_i >= right:\n                break\n            move_right[i] = right\n            i = next_i\n    for y1 in [y-R,y+R]:\n        if not y1 in Y_to_X:\n            continue\n        arr, move_right = Y_to_X[y1]\n        left = bisect.bisect_left(arr, x-R+1)\n        right = bisect.bisect_right(arr, x+R-1)\n        # [left, right) が辺で結べるターゲット\n        answer += right - left # 辺の個数\n        i = left\n        while i < right:\n            x1 = arr[i]\n            if (x1,y1) not in equiv_class:\n                equiv_class.add((x1,y1))\n                task.append((x1,y1))\n            # なるべく、既に居ない元は見ないで済ませるように\n            next_i = move_right[i]\n            if next_i >= right:\n                break\n            move_right[i] = right\n            i = next_i\n\nanswer //= 2 # 両方向から辺を見た\nprint(answer)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 15095