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;
string second[15];
char t[55555], ss[55555];
int p[12][55555], LEN[12], z[55555], z2[55555];
bool can[2222][2222];
int add2[11][2222], add[2222][2222];
int sum[11][2222], L[11], R[11], cur[11], pos[11], n, len;
int main() {
  scanf("%s", &t);
  len = strlen(t);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", &ss);
    second[i] = ss;
    LEN[i] = second[i].size();
    scanf("%d%d", &L[i], &R[i]);
  }
  memset(can, 1, sizeof(can));
  for (int i = 0; i < len; ++i) {
    memset(add2, 0, sizeof(add2));
    z[i] = 0;
    int j = i;
    for (int k = i + 1; k < len; ++k) {
      z[k] = 0;
      if (j + z[j] > k) {
        z[k] = min(j + z[j] - k, z[i + k - j]);
      }
      while (k + z[k] < len && t[k + z[k]] == t[i + z[k]]) ++z[k];
      if (j + z[j] < k + z[k]) j = k;
      if (z[k] > 0) {
        ++add[k][k + z[k] - 1];
      }
    }
    for (int j = 0; j < n; ++j) {
      int k = -1;
      for (int q = 0; q < LEN[j]; ++q) {
        z2[q] = 0;
        if (k != -1 && k + z2[k] > q) {
          z2[q] = min(k + z2[k] - q, z[i + q - k]);
        }
        while (q + z2[q] < LEN[j] && i + z2[q] < len &&
               t[i + z2[q]] == second[j][q + z2[q]])
          ++z2[q];
        while (k == -1 || q + z2[q] > k + z2[k]) k = q;
        if (z2[q] > 0) ++add2[j][i + z2[q] - 1];
      }
    }
    memset(cur, 0, sizeof(cur));
    for (int k = len - 1; k >= i; --k) {
      bool ok = true;
      for (int j = 0; j < n; ++j) {
        cur[j] += add2[j][k];
        if (cur[j] < L[j] || cur[j] > R[j]) ok = false;
      }
      can[i][k] = ok;
    }
  }
  int res = 0;
  for (int i = 0; i < len; ++i) {
    int second = 0;
    for (int j = len - 1; j >= i; --j) {
      second += add[i][j];
      if (second == 0 && can[i][j]) {
        ++res;
      }
    }
  }
  cout << res << endl;
  return 0;
}
