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
# AOJ 2608: Minus One
# Python3 2018.6.30 bal4u

import heapq
MAX = 0x7fffffff

def dijkstra(dist, start):
	Q = []
	dist[start] = 0
	heapq.heappush(Q, (0, start))
	while Q:
		t, s = heapq.heappop(Q)
		if dist[s] < t: continue
		for e in to[s]:
			nt = t + 1
			if dist[e] > nt:
				dist[e] = nt
				heapq.heappush(Q, (nt, e))

N, M, s, t = map(int, input().split())
s, t = s-1, t-1
to = [[] for i in range(N)]
for i in range(M):
	x, y = map(int, input().split())
	x, y = x-1, y-1
	to[x].append(y)
	to[y].append(x)
smin = [MAX]*N
dijkstra(smin, s)
tmin = [MAX]*N
dijkstra(tmin, t)
scnt = [0]*N
tcnt = [0]*N
for i in range(N):
	if smin[i] < N: scnt[smin[i]] += 1
	if tmin[i] < N: tcnt[tmin[i]] += 1
ans = 0
s = smin[t]-1
for i in range(s): ans += scnt[i]*tcnt[s-1-i]
print(ans)
