Little town Nsk consists of *n* junctions connected by *m* bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.

In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction *s* to work located near junction *t*. Thus, he wants to build a new road in such a way that the distance between these two junctions won't decrease. 

You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between *s* and *t* won't decrease.
The firt line of the input contains integers *n*, *m*, *s* and *t* (2 ≤ *n* ≤ 1000, 1 ≤ *m* ≤ 1000, 1 ≤ *s*, *t* ≤ *n*, *s* ≠ *t*) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The *i*-th of the following *m* lines contains two integers *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub> (1 ≤ *u*<sub class="lower-index">*i*</sub>, *v*<sub class="lower-index">*i*</sub> ≤ *n*, *u*<sub class="lower-index">*i*</sub> ≠ *v*<sub class="lower-index">*i*</sub>), meaning that this road connects junctions *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub> directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.
none
5 4 1 5
1 2
2 3
3 4
4 5
0
5 4 3 5
1 2
2 3
3 4
4 5
5
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
3
2 1 2 1
1 2
0
3 2 2 3
1 2
2 3
1
3 2 1 3
1 2
2 3
0
3 3 2 3
1 2
2 3
1 3
0
I'll try to code that. Here is a possible solution in Python:

```python
# Read the input
n, m, s, t = map(int, input().split())
s -= 1 # Adjust the indices to start from 0
t -= 1
roads = [] # List of existing roads
for _ in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    roads.append((u, v))

# Use BFS to find the shortest distance from s to t
dist = [-1] * n # Distance array
dist[s] = 0 # Distance from s to s is 0
queue = [s] # Queue for BFS
while queue:
    u = queue.pop(0) # Dequeue the first element
    for v in range(n): # Loop through all possible junctions
        if v != u and (u, v) not in roads and (v, u) not in roads and dist[v] == -1: # If there is no road between u and v and v is not visited
            dist[v] = dist[u] + 1 # Update the distance of v
            queue.append(v) # Enqueue v for BFS

# Count the number of pairs that won't decrease the distance from s to t
ans = 0 # Answer variable
for u in range(n): # Loop through all possible junctions for the first end of the new road
    for v in range(u + 1, n): # Loop through all possible junctions for the second end of the new road
        if (u, v) not in roads and (v, u) not in roads: # If there is no road between u and v
            if dist[u] + dist[v] + 1 >= dist[t]: # If adding a road between u and v won't decrease the distance from s to t
                ans += 1 # Increment the answer

# Print the answer
print(ans)
```