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 INF = 1000000007;
const int N = 2020;
struct trie {
  trie *c[26];
  int cnt;
} * root, *nil;
string s1, s2, s3;
int n, l, r, f[N][N], ans;
trie *new_node() {
  trie *node = new (trie);
  for (int i = 0; i < 26; i++) node->c[i] = nil;
  node->cnt = 0;
  return node;
}
void deleteTree(trie *root) {
  for (int i = 0; i < 26; i++)
    if (root->c[i] != nil) {
      deleteTree(root->c[i]);
      delete (root->c[i]);
    }
}
int main() {
  getline(cin, s1);
  scanf("%d\n", &n);
  nil = new (trie);
  root = new_node();
  for (int i = 1; i <= n; i++) {
    deleteTree(root);
    for (int t = 0; t < 26; t++) root->c[t] = nil;
    cin >> s2 >> l >> r;
    for (int u = 0; u < s2.size(); u++) {
      trie *path = root;
      for (int v = u; v < s2.size(); v++) {
        int q = s2[v] - 'a';
        if (path->c[q] == nil) {
          trie *nw = new_node();
          path->c[q] = nw;
        }
        path = path->c[q];
        path->cnt++;
      }
    }
    for (int u = 0; u < s1.size(); u++) {
      trie *path = root;
      for (int v = u; v < s1.size(); v++) {
        int q = s1[v] - 'a';
        if (path->c[q] == nil) {
          if (l == 0)
            for (int t = v; t < s1.size(); t++) f[u][t]++;
          break;
        }
        path = path->c[q];
        if (path->cnt >= l && path->cnt <= r) f[u][v]++;
      }
    }
  }
  deleteTree(root);
  for (int t = 0; t < 26; t++) root->c[t] = nil;
  for (int i = 0; i < s1.size(); i++) {
    trie *path = root;
    for (int j = i; j < s1.size(); j++) {
      int q = s1[j] - 'a';
      if (path->c[q] == nil) {
        trie *nw = new_node();
        path->c[q] = nw;
      }
      path = path->c[q];
      if (f[i][j] == n && path->cnt == 0) ans++;
      path->cnt++;
    }
  }
  cout << ans << endl;
}
