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
#include <bits/stdc++.h>
template<typename T>
struct Edge {
    int from, to;
    T cost;
    Edge(int from, int to, T cost) {
        this->from = from;
        this->to = to;
        this->cost = cost;
    }
};
bool operator == (Edge<int> e1, Edge<int> e2) {
    return e1.from == e2.from &&
            e1.to == e2.to &&
            e1.cost == e2.cost;
}
template<typename T>
using Edges = std::vector<Edge<T>>;
template<typename T>
using Graph = std::vector<Edges<T>>;
template <typename T>
std::vector<T> dijkstra(const Graph<T> &g, int s) {
    int sz = (int)(g.size());
    std::vector<T> dist(sz, std::numeric_limits<T>::max());
    std::priority_queue<std::pair<T, int>, std::vector<std::pair<T, int>>, std::greater<std::pair<T, int>>> pq;
    dist[s] = 0;
    pq.push({0, s});
    while(!pq.empty()) {
        int d = pq.top().first, now = pq.top().second;
        pq.pop();
        if(d > dist[now]) continue;
        for(int i=0;i<(int)(g[now].size());++i) {
            int next = g[now][i].to, cost = g[now][i].cost;
            if(dist[next] > dist[now] + cost) {
                dist[next] = dist[now] + cost;
                pq.push({cost, next});
            }
        }
    }
    return dist;
}
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using vi = vector<ll>;
using vvi = vector<vector<ll>>;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
const double eps = 1e-10;
const ll MOD = 1000000007;
const ll INF = 1000000000;
const ll LINF = 1ll<<50;
template<typename T>
void printv(const vector<T>& s) {
  for(ll i=0;i<(ll)(s.size());++i) {
    cout << s[i];
    if(i == (ll)(s.size())-1) cout << endl;
    else cout << " ";
  }
}
signed main() {
  cin.tie(0);
  cout << fixed << setprecision(10);
  ll n, m, s, t; cin >> n >> m >> s >> t;
  s--; t--;
  Graph<ll> g(n);
  vector<P> eds;
  for(ll i=0;i<m;++i) {
    ll u, v; cin >> u >> v;
    u--; v--;
    g[u].push_back(Edge<ll>(u, v, 1));
    g[v].push_back(Edge<ll>(v, u, 1));
    eds.push_back({u, v});
  }
  vector<ll> froms = dijkstra(g, s), fromt = dijkstra(g, t);
  ll d = froms[t];
  map<ll, ll> cnts, cntt;
  for(ll i=0;i<n;++i) {
    cnts[froms[i]]++;
  }
  for(ll i=0;i<n;++i) {
    cntt[fromt[i]]++;
  }
  ll ans = 0;
  for(ll i=0;i<=d-2;++i) {
    ans += cnts[i] * cntt[d-2 - i];
  }
  cout << ans << endl;
}

