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;
const int maxn = 5e4 + 100;
int dp[2 * maxn][15];
struct state {
  int len;
  int link;
  int next[30];
} st[2 * maxn];
int sz = 0;
void init() {
  sz++;
  st[0].len = 0;
  st[0].link = -1;
  st[0].len = 0;
}
int last = 0;
void extend(int c, int id) {
  int cur = sz++;
  st[cur].len = st[last].len + 1;
  dp[cur][id]++;
  int p = last;
  while (p != -1 && !st[p].next[c]) {
    st[p].next[c] = cur;
    p = st[p].link;
  }
  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] = st[q];
      st[clone].len = st[p].len + 1;
      while (p != -1 && st[p].next[c] == q) {
        st[p].next[c] = clone;
        p = st[p].link;
      }
      st[cur].link = st[q].link = clone;
    }
  }
  last = cur;
}
int deg[2 * maxn];
int m;
int l[15], r[15];
long long count() {
  long long ans = 0;
  queue<int> q;
  for (int i = 1; i < sz; i++) deg[st[i].link]++;
  for (int i = 1; i < sz; i++) {
    if (deg[i] == 0) q.push(i);
  }
  while (q.size()) {
    int at = q.front();
    q.pop();
    for (int i = 0; i <= m; i++) {
      dp[st[at].link][i] += dp[at][i];
    }
    deg[st[at].link]--;
    if (deg[st[at].link] == 0) q.push(st[at].link);
    if (dp[at][0] == 0) continue;
    bool ok = 1;
    for (int i = 1; i <= m; i++) {
      ok &= (l[i] <= dp[at][i] && dp[at][i] <= r[i]);
    }
    if (ok) ans += st[at].len - st[st[at].link].len;
  }
  return ans;
}
char s[maxn];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  init();
  for (int i = 0; i < n; i++) {
    extend(s[i] - 'a', 0);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s %d %d", s, &l[i], &r[i]);
    n = strlen(s);
    extend(26, i);
    for (int j = 0; j < n; j++) {
      extend(s[j] - 'a', i);
    }
  }
  printf("%lld\n", count());
  return 0;
}
