Problem: There are n players numbered from 0 to n-1 with ranks. The i-th player has rank i.

Players can form teams: the team should consist of three players and no pair of players in the team should have a conflict. The rank of the team is calculated using the following algorithm: let i, j, k be the ranks of players in the team and i < j < k, then the rank of the team is equal to A ⋅ i + B ⋅ j + C ⋅ k.

You are given information about the pairs of players who have a conflict. Calculate the total sum of ranks over all possible valid teams modulo 2^{64}.

Input

The first line contains two space-separated integers n and m (3 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ m ≤ 2 ⋅ 10^5) — the number of players and the number of conflicting pairs.

The second line contains three space-separated integers A, B and C (1 ≤ A, B, C ≤ 10^6) — coefficients for team rank calculation.

Each of the next m lines contains two space-separated integers u_i and v_i (0 ≤ u_i, v_i < n, u_i ≠ v_i) — pair of conflicting players.

It's guaranteed that each unordered pair of players appears in the input file no more than once.

Output

Print single integer — the total sum of ranks over all possible teams modulo 2^{64}.

Examples

Input

4 0
2 3 4


Output

64


Input

4 1
2 3 4
1 0


Output

38


Input

6 4
1 5 3
0 3
3 5
5 4
4 3


Output

164

Note

In the first example all 4 teams are valid, i.e. triples: {0, 1, 2}, {0, 1, 3}, {0, 2, 3} {1, 2, 3}.

In the second example teams are following: {0, 2, 3}, {1, 2, 3}.

In the third example teams are following: {0, 1, 2}, {0, 1, 4}, {0, 1, 5}, {0, 2, 4}, {0, 2, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 2e5 + 5;
const int magic = 300;
int n, m;
int a, b, c;
vector<int> adj[maxn];
unsigned long long f[maxn];
unsigned long long g[maxn];
unsigned long long h[maxn];
void calc(unsigned long long f[]) {
  map<pair<int, int>, int> hs;
  for (int u = (0); u < (n); ++u)
    for (int v : adj[u]) hs[make_pair(u, v)] = 1;
  unsigned long long sum = 0;
  for (int u = (0); u < (n); ++u) {
    f[u] = sum;
    vector<int> ver;
    for (int v : adj[u])
      if (v < u) {
        ver.push_back(v);
        f[u] += u - 1;
        f[u] -= upper_bound((adj[v]).begin(), (adj[v]).end(), u - 1) -
                adj[v].begin();
      }
    f[u] -= (unsigned long long)int((ver).size()) * (int((ver).size()) - 1) / 2;
    if (int((ver).size()) < magic) {
      for (int i = (0); i < (int((ver).size())); ++i)
        for (int j = (i + 1); j < (int((ver).size())); ++j) {
          if (hs[make_pair(ver[i], ver[j])]) {
            f[u]++;
          }
        }
    } else {
      static int flag[maxn];
      for (int v : ver) {
        for (int w : adj[v]) {
          if (w > v) break;
          f[u] += flag[w];
        }
        flag[v] = 1;
      }
      for (int v : ver) {
        flag[v] = 0;
      }
    }
    sum += int((ver).size());
    f[u] = (unsigned long long)u * (u - 1) / 2 - f[u];
  }
}
void chemthan() {
  cin >> n >> m >> a >> b >> c;
  for (int i = (0); i < (m); ++i) {
    int u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  for (int u = (0); u < (n); ++u) sort((adj[u]).begin(), (adj[u]).end());
  calc(f);
  for (int u = 0; u + u <= n - 1; u++) {
    swap(adj[u], adj[n - u - 1]);
  }
  for (int u = (0); u < (n); ++u) {
    for (int& v : adj[u]) v = n - v - 1;
    reverse((adj[u]).begin(), (adj[u]).end());
  }
  calc(g);
  for (int u = 0; u + u <= n - 1; u++) {
    swap(g[u], g[n - u - 1]);
    swap(adj[u], adj[n - u - 1]);
  }
  for (int u = (0); u < (n); ++u) {
    for (int& v : adj[u]) v = n - v - 1;
    reverse((adj[u]).begin(), (adj[u]).end());
  }
  map<pair<int, int>, int> hs;
  for (int u = (0); u < (n); ++u)
    for (int v : adj[u]) hs[make_pair(u, v)] = 1;
  for (int u = (0); u < (n); ++u) {
    h[u] = m - int((adj[u]).size());
    vector<int> ver;
    for (int v : adj[u]) {
      ver.push_back(v);
      h[u] += n - 2;
      h[u] -= int((adj[v]).size()) - 1;
    }
    h[u] -= (unsigned long long)int((ver).size()) * (int((ver).size()) - 1) / 2;
    if (int((ver).size()) < magic) {
      for (int i = (0); i < (int((ver).size())); ++i)
        for (int j = (i + 1); j < (int((ver).size())); ++j) {
          if (hs[make_pair(ver[i], ver[j])]) {
            h[u]++;
          }
        }
    } else {
      static int flag[maxn];
      for (int v : ver) {
        for (int w : adj[v]) {
          if (w > v) break;
          h[u] += flag[w];
        }
        flag[v] = 1;
      }
      for (int v : ver) {
        flag[v] = 0;
      }
    }
    h[u] = (unsigned long long)(n - 1) * (n - 2) / 2 - h[u];
  }
  unsigned long long res = 0;
  for (int u = (0); u < (n); ++u) {
    res += c * f[u] * u;
    res += a * g[u] * u;
    res += b * (h[u] - f[u] - g[u]) * u;
  }
  cout << res << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}