p01720 Minus One
Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2
```python
from collections import deque, defaultdict
n, m, s, t = map(int, input().split())
s -= 1
t -= 1
edges = [[] for _ in range(n)]
for _ in range(m):
  x, y = map(int, input().split())
  x -= 1
  y -= 1
  edges[x].append(y)
  edges[y].append(x)

def dist_from(s):
  INF = 10 ** 20
  dist = [INF] * n
  dist[s] = 0
  que = deque()
  que.append((0, s))
  while que:
    score, node = que.popleft()
    for to in edges[node]:
      if dist[to] > score + 1:
        dist[to] = score + 1
        que.append((score + 1, to))
  return dist

dist_from_s = dist_from(s)
dic1 = {}
for i, d in enumerate(dist_from_s):
  if d in dic1:
    dic1[d].add(i)
  else:
    dic1[d] = {i}

dist_from_t = dist_from(t)
dic2 = {}
for i, d in enumerate(dist_from_t):
  if d in dic2:
    dic2[d].add(i)
  else:
    dic2[d] = {i}

st_dist = dist_from_s[t]
ans = 0
for key in dic1:
  another_key = st_dist - key - 2
  if another_key in dic2:
    add = len(dic1[key]) * len(dic2[another_key])
    for u in dic1[key]:
      for to in edges[u]:
        if to in dic2[another_key]:
          add -= 1
    ans += add
print(ans)


```