Smart Beaver recently got interested in a new word game. The point is as follows: count the number of distinct good substrings of some string s. To determine if a string is good or not the game uses rules. Overall there are n rules. Each rule is described by a group of three (p, l, r), where p is a string and l and r (l ≤ r) are integers. We’ll say that string t complies with rule (p, l, r), if the number of occurrences of string t in string p lies between l and r, inclusive. For example, string "ab", complies with rules ("ab", 1, 2) and ("aab", 0, 1), but does not comply with rules ("cd", 1, 2) and ("abab", 0, 1).

A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (|s| is a length of s) is string slsl + 1... sr.

Consider a number of occurrences  of string t in string p as a number of pairs of integers l, r (1 ≤ l ≤ r ≤ |p|) such that p[l... r] = t.

We’ll say that string t is good if it complies with all n rules. Smart Beaver asks you to help him to write a program that can calculate the number of distinct good substrings of string s. Two substrings s[x... y] and s[z... w] are cosidered to be distinct iff s[x... y] ≠ s[z... w].

Input

The first line contains string s. The second line contains integer n. Next n lines contain the rules, one per line. Each of these lines contains a string and two integers pi, li, ri, separated by single spaces (0 ≤ li ≤ ri ≤ |pi|). It is guaranteed that all the given strings are non-empty and only contain lowercase English letters.

The input limits for scoring 30 points are (subproblem G1): 

  * 0 ≤ n ≤ 10. 
  * The length of string s and the maximum length of string p is  ≤ 200. 



The input limits for scoring 70 points are (subproblems G1+G2): 

  * 0 ≤ n ≤ 10. 
  * The length of string s and the maximum length of string p is  ≤ 2000. 



The input limits for scoring 100 points are (subproblems G1+G2+G3): 

  * 0 ≤ n ≤ 10. 
  * The length of string s and the maximum length of string p is  ≤ 50000. 

Output

Print a single integer — the number of good substrings of string s.

Examples

Input

aaab
2
aa 0 0
aab 1 1


Output

3


Input

ltntlnen
3
n 0 0
ttlneenl 1 4
lelllt 1 1


Output

2


Input

a
0


Output

1

Note

There are three good substrings in the first sample test: «aab», «ab» and «b».

In the second test only substrings «e» and «t» are good.
#include <bits/stdc++.h>
using namespace std;
struct state {
  int len, link;
  map<int, int> next;
};
const int MAXLEN = 50000 * 30;
state st[MAXLEN];
int sz, last;
void sa_init() {
  sz = last = 0;
  st[0].len = 0;
  st[0].link = -1;
  ++sz;
}
void sa_extend(char c) {
  int cur = sz++;
  st[cur].len = st[last].len + 1;
  int p;
  for (p = last; p != -1 && !st[p].next.count(c); p = st[p].link)
    st[p].next[c] = cur;
  if (p == -1)
    st[cur].link = 0;
  else {
    int q = st[p].next[c];
    if (st[p].len + 1 == st[q].len)
      st[cur].link = q;
    else {
      int clone = sz++;
      st[clone].len = st[p].len + 1;
      st[clone].next = st[q].next;
      st[clone].link = st[q].link;
      for (; p != -1 && st[p].next[c] == q; p = st[p].link)
        st[p].next[c] = clone;
      st[q].link = st[cur].link = clone;
    }
  }
  last = cur;
}
string s;
int n, tot, l[22], r[22];
long long ans;
int dp[MAXLEN][12];
bool vis[MAXLEN];
void fun(int u) {
  if (vis[u]) {
    return;
  }
  vis[u] = true;
  for (int i = 0; i <= n; ++i) {
    if (st[u].next.count(26 + i)) {
      dp[u][i] = 1;
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (st[u].next.count(i)) {
      int v = st[u].next[i];
      fun(v);
      for (int j = 0; j <= n; ++j) {
        dp[u][j] += dp[v][j];
      }
    }
  }
  if (u && dp[u][0]) {
    bool flag = true;
    for (int i = 1; i <= n && flag; ++i) {
      flag = l[i] <= dp[u][i] && dp[u][i] <= r[i];
    }
    if (flag) {
      ans += st[u].len - st[st[u].link].len;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  n = 0;
  sa_init();
  for (int i(0), _n(((int)(s.size()))); i < _n; ++i) sa_extend(s[i] - 'a');
  sa_extend(26);
  cin >> n;
  for (int i(0), _n(n); i < _n; ++i) {
    cin >> s;
    cin >> l[i + 1] >> r[i + 1];
    for (int j(0), _n(((int)(s.size()))); j < _n; ++j) sa_extend(s[j] - 'a');
    sa_extend(27 + i);
  }
  fun(0);
  cout << ans;
}
