Problem: Consider a string s of n lowercase English letters. Let t_i be the string obtained by replacing the i-th character of s with an asterisk character *. For example, when s = abc, we have t_1 = \tt{*bc}, t_2 = \tt{a*c}, and t_3 = \tt{ab*}.

Given a string s, count the number of distinct strings of lowercase English letters and asterisks that occur as a substring of at least one string in the set \\{s, t_1, …, t_n \}. The empty string should be counted.

Note that *'s are just characters and do not play any special role as in, for example, regex matching.

Input

The only line contains a string s of n lowercase English letters (1 ≤ n ≤ 10^5).

Output

Print a single integer — the number of distinct strings of s, t_1, …, t_n.

Example

Input


abc


Output


15

Note

For the sample case, the distinct substrings are (empty string), a, b, c, *, ab, a*, bc, b*, *b, *c, abc, ab*, a*c, *bc.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int gi() {
  char cc = getchar();
  int cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const int inf = 1e9;
const int Inf = 1e8;
const int N = 2e5 + 5;
int n, rt[N];
long long ans[N], Ans;
char sf[N], us[N];
struct SuffixTree {
  struct SufTree {
    int lef, len, lk, ch[30];
  } t[N];
  int idx, M, cnt, Dep[N], dfn[N], Id[N], s[N], dep[N];
  int fd[N], ind[N], f[N][20];
  long long F;
  int node(int l, int len) {
    ++cnt, t[cnt].lef = l, t[cnt].len = len;
    return cnt;
  }
  void build(char *a, int l) {
    t[0].len = inf, M = l, a[++l] = 'z' + 1;
    for (int i = (1); i <= (l); ++i) s[i] = a[i] - 'a' + 1;
    cnt = 1;
    int rem = 0, m = 0, bef = 1;
    for (int o = (1); o <= (l); ++o) {
      int u = s[o], lst = 1;
      ++m, ++rem;
      while (rem) {
        while (rem > t[t[bef].ch[s[m - rem + 1]]].len)
          rem -= t[bef = t[bef].ch[s[m - rem + 1]]].len;
        int &d = t[bef].ch[s[m - rem + 1]], c = s[t[d].lef + rem - 1];
        if (!d || u == c) {
          t[lst].lk = bef, lst = bef;
          if (!d)
            d = node(m - rem + 1, inf);
          else
            break;
        } else {
          int np = node(t[d].lef, rem - 1);
          t[d].lef += (rem - 1), t[d].len -= (rem - 1),
              t[np].ch[c] = d, t[np].ch[u] = node(m, inf), t[lst].lk = d = np,
              lst = np;
        }
        (bef == 1) ? --rem : bef = t[bef].lk;
      }
    }
  }
  void clear() {
    for (int i = (1); i <= (cnt); ++i) {
      Dep[i] = dep[i] = ind[i] = Id[i] = dfn[i] = 0;
      t[i].lef = t[i].len = t[i].lk = 0;
      for (int j = (0); j <= (17); ++j) f[i][j] = 0;
      for (int j = (0); j <= (27); ++j) t[i].ch[j] = 0;
    }
    idx = cnt = F = 0;
  }
  void DFS(int x, int fa) {
    f[x][0] = fa, dfn[x] = ++idx, Dep[x] = Dep[fa] + 1, Id[idx] = x;
    for (int i = (1); i <= (17); ++i) f[x][i] = f[f[x][i - 1]][i - 1];
    if (t[x].len >= Inf) {
      t[x].len = M - t[x].lef + 1, dep[x] = dep[fa] + t[x].len;
      F += t[x].len, fd[idx] = dep[x], ind[M - dep[x] + 1] = x;
      return;
    }
    dep[x] = dep[fa] + t[x].len, F += t[x].len, fd[idx] = dep[x];
    for (int i = (1); i <= (27); ++i)
      if (t[x].ch[i]) DFS(t[x].ch[i], x);
  }
  int LCA(int x, int y) {
    if (Dep[x] < Dep[y]) swap(x, y);
    for (int i = (17); i >= (0); --i)
      if (Dep[f[x][i]] >= Dep[y]) x = f[x][i];
    for (int i = (17); i >= (0); --i)
      if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
    return (x == y) ? x : f[x][0];
  }
  int lca(int l1, int l2) { return dep[LCA(Id[l1], Id[l2])]; }
} T1, uT;
set<int> S[N];
set<int>::iterator it;
void merge(int x, int y) {
  if (S[rt[x]].empty() && S[rt[y]].empty()) return;
  if (S[rt[x]].size() < S[rt[y]].size()) swap(rt[x], rt[y]);
  for (int v : S[rt[y]]) {
    it = S[rt[x]].lower_bound(v);
    int nxt = 0, pre = 0;
    if (it != S[rt[x]].end()) nxt = (*it), ans[rt[x]] -= uT.lca(v, nxt);
    if (it != S[rt[x]].begin()) --it, pre = (*it), ans[rt[x]] -= uT.lca(v, pre);
    if (pre && nxt) ans[rt[x]] += uT.lca(pre, nxt);
    S[rt[x]].insert(v), ans[rt[x]] += uT.fd[v];
  }
}
void solve(int x) {
  int flag = 0, d = T1.dep[x];
  for (int i = (1); i <= (27); ++i)
    if (T1.t[x].ch[i]) solve(T1.t[x].ch[i]), flag = 1, merge(x, T1.t[x].ch[i]);
  if (!flag && (d <= n - 2) && (d > 0))
    S[x].insert(uT.dfn[uT.ind[d + 2]]), ans[x] = n - d - 1;
  Ans += 1ll * ans[rt[x]] * T1.t[x].len;
}
void output(int x) {
  for (int i = (1); i <= (27); ++i)
    if (uT.t[x].ch[i]) output(uT.t[x].ch[i]);
}
signed main() {
  scanf("%s", sf + 1), n = strlen(sf + 1), Ans = 2;
  for (int i = (2); i <= (n); ++i) us[i - 1] = sf[i];
  uT.build(us, n - 1), uT.DFS(1, 1), Ans += uT.F, uT.clear();
  for (int i = (1); i <= (n - 1); ++i) us[i] = sf[i];
  uT.build(us, n - 1), uT.DFS(1, 1), Ans += uT.F, uT.clear();
  for (int i = (1); i <= (n); ++i) us[i] = sf[i];
  uT.build(us, n), uT.DFS(1, 1), Ans += uT.F;
  for (int i = (1); i <= (n); ++i) us[n - i + 1] = sf[i];
  T1.build(us, n), T1.DFS(1, 1);
  for (int i = (1); i <= (T1.cnt); ++i) rt[i] = i;
  solve(1), cout << Ans << endl;
  return 0;
}