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;
char s[10001];
char t[20][10001];
int tl[20];
int l[10001];
int r[10001];
int n, len;
int z[10001];
char b[10001];
char good[4005][4005];
const unsigned long long mod = 123526111;
const unsigned long long mod2 = 1234567891;
const unsigned long long alpha = 31;
unsigned long long h[5005];
unsigned long long h2[5005];
unsigned long long apow[5005];
unsigned long long apow2[5005];
set<pair<unsigned long long, unsigned long long> > ht;
int res = 0;
void addh(int l, int r) {
  unsigned long long hl = h[l] * apow[r - l];
  hl %= mod;
  unsigned long long ch = (h[r] - hl + mod) % mod;
  unsigned long long hl2 = h2[l] * apow2[r - l];
  hl2 %= mod2;
  unsigned long long ch2 = (h2[r] - hl2 + mod2) % mod2;
  if (ht.find(make_pair(ch, ch2)) == ht.end()) {
    ht.insert(make_pair(ch, ch2));
    ++res;
  }
}
inline void calc(int n) {
  for (int i = 0; i <= n; ++i) z[i] = 0;
  int l = 0, r = 0;
  for (int i = 1; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && b[z[i]] == b[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) {
      l = i;
      r = i + z[i] - 1;
    }
  }
}
int sum[100001];
void solve(const int lb) {
  int ret = 0;
  for (int i = lb + 1; i <= len; ++i) good[lb][i] = 1;
  int bl = 0;
  for (int i = lb; i < len; ++i) b[bl++] = s[i];
  b[bl++] = '#';
  int bs = bl;
  for (int ii = 0; ii < n; ++ii) {
    bl = bs;
    for (int i = 0; i < tl[ii]; ++i) b[bl++] = t[ii][i];
    calc(bl);
    for (int i = 0; i <= bs; ++i) sum[i] = 0;
    for (int i = bs; i < bl; ++i) {
      sum[z[i]]++;
    }
    int cs = 0;
    for (int i = bs; i >= 0; --i) {
      cs += sum[i];
      sum[i] = cs;
    }
    for (int i = 1; i < bs; ++i)
      if (!(l[ii] <= sum[i] && sum[i] <= r[ii])) {
        good[lb][lb + i] = 0;
      }
  }
}
int main() {
  gets(s);
  len = strlen(s);
  for (int i = 0; i < len; ++i) {
    h[i + 1] = h[i] * alpha + (s[i] - 'a' + 1);
    h[i + 1] %= mod;
    h2[i + 1] = h2[i] * alpha + (s[i] - 'a' + 1);
    h2[i + 1] %= mod2;
  }
  apow[0] = apow2[0] = 1;
  for (int i = 0; i <= len; ++i) {
    apow[i + 1] = (apow[i] * alpha) % mod;
    apow2[i + 1] = (apow2[i] * alpha) % mod2;
  }
  scanf("%d\n", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s%d%d\n", t[i], l + i, r + i);
    tl[i] = strlen(t[i]);
  }
  for (int i = 0; i < len; ++i) solve(i);
  n = len;
  for (int len = 1; len <= n; ++len) {
    for (int i = 0; i + len <= n; ++i)
      if (good[i][i + len]) {
        addh(i, i + len);
      }
    ht.clear();
  }
  cout << res << '\n';
}
