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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  if (!x) return;
  recursive_print(x / 10);
  putc(x % 10 ^ 48);
}
template <typename T>
void print(T x) {
  if (!x) putc('0');
  if (x < 0) putc('-'), x = ~x + 1;
  recursive_print(x);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MAXN = 2e5;
int n, m, deg[MAXN + 5];
unsigned long long a, b, c;
unsigned long long sum(unsigned long long l, unsigned long long r) {
  return (l + r) * (r - l + 1) / 2ull;
}
vector<int> g1[MAXN + 5], g2[MAXN + 5], g[MAXN + 5];
vector<unsigned long long> s1[MAXN + 5], s2[MAXN + 5];
int vis[MAXN + 5];
int main() {
  scanf("%d%d%llu%llu%llu", &n, &m, &a, &b, &c);
  for (int i = 1, u, v; i <= m; i++) {
    scanf("%d%d", &u, &v);
    if (u > v) u ^= v ^= u ^= v;
    g1[v].push_back(u);
    g2[u].push_back(v);
    deg[u]++;
    deg[v]++;
  }
  unsigned long long ans = 0;
  for (int i = 0; i < n; i++) {
    sort(g1[i].begin(), g1[i].end());
    s1[i].resize(g1[i].size());
    for (int j = 0; j < g1[i].size(); j++) {
      if (j == 0)
        s1[i][j] = g1[i][j];
      else
        s1[i][j] = s1[i][j - 1] + g1[i][j];
    }
    sort(g2[i].begin(), g2[i].end());
    s2[i].resize(g2[i].size());
    for (int j = 0; j < g2[i].size(); j++) {
      if (j == 0)
        s2[i][j] = g2[i][j];
      else
        s2[i][j] = s2[i][j - 1] + g2[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    unsigned long long cnt1 = i - g1[i].size(), cnt2 = n - i - 1 - g2[i].size();
    unsigned long long sum1 =
        sum(0, i - 1) - ((g1[i].empty()) ? 0 : s1[i].back());
    unsigned long long sum2 =
        sum(i + 1, n - 1) - ((g2[i].empty()) ? 0 : s2[i].back());
    ans += cnt1 * cnt2 * i * b + cnt1 * sum2 * c + cnt2 * sum1 * a;
  }
  for (int i = 0; i < n; i++)
    for (int j : g2[i]) {
      int pos1 = lower_bound(g1[j].begin(), g1[j].end(), i) - g1[j].begin();
      int pos2 = lower_bound(g2[i].begin(), g2[i].end(), j) - g2[i].begin();
      unsigned long long cnt = g1[j].size() - 1 - pos1 + pos2;
      unsigned long long rem = j - i - 1 - cnt;
      unsigned long long s = sum(i + 1, j - 1) - s1[j].back() + s1[j][pos1] -
                             ((!pos2) ? 0 : s2[i][pos2 - 1]);
      ans -= s * b + rem * i * a + rem * j * c;
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < g1[i].size(); j++) {
      int x = g1[i][j];
      if (deg[i] < deg[x])
        g[i].push_back(x);
      else
        g[x].push_back(i);
    }
  memset(vis, -1, sizeof(vis));
  for (int i = 0; i < n; i++) {
    for (int j : g[i]) vis[j] = i;
    for (int j : g[i])
      for (int k : g[j]) {
        if (vis[k] == i) {
          unsigned long long tmp[3] = {0};
          tmp[0] = i;
          tmp[1] = j;
          tmp[2] = k;
          sort(tmp, tmp + 3);
          ans -= tmp[0] * a + tmp[1] * b + tmp[2] * c;
        }
      }
  }
  printf("%llu\n", ans);
  return 0;
}