#include <bits/stdc++.h>
using namespace std;
char s[100500];
char c[100500];
char st[100500];
int pf[100500];
bool isEnter[50500];
int len;
int lft[11][50500];
int rght[11][50500];
void calcPf() {
  int len = strlen(c);
  int sWin = 0, eWin = 0;
  for (int i = 1; i < len; i++) {
    int curpf = 0;
    if (eWin > i) {
      curpf = pf[i - sWin];
    }
    curpf = min(curpf, eWin - i);
    curpf = max(0, curpf);
    while (c[i + curpf] == c[curpf]) curpf++;
    if (i + curpf > eWin) {
      eWin = i + curpf;
      sWin = i;
    }
    pf[i] = curpf;
  }
}
int cnt[50500];
bool good[2005][2005];
void castRule(int rule) {
  memset(isEnter, false, sizeof isEnter);
  int a, b;
  scanf("%s %d %d", st, &a, &b);
  int curLen = strlen(st);
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) c[j - i] = s[j];
    c[len - i] = '$';
    for (int j = 0; j < curLen; j++) c[j + len - i + 1] = st[j];
    c[curLen + len - i + 1] = 0;
    calcPf();
    for (int j = 0; j < len; j++) cnt[j] = 0;
    for (int j = 0; j < curLen; j++) cnt[pf[j + len - i + 1]]++;
    int curCnt = curLen;
    for (int j = i; j <= len; j++) {
      if (curCnt < a || curCnt > b) good[i][j] = false;
      curCnt -= cnt[j - i];
    }
  }
}
int nxt[2000500][26];
bool isFin[2000500];
int lst = 1;
int addStr(int l, int r) {
  int curpos = 0;
  for (int i = l; i < r; i++) {
    char ch = s[i] - 'a';
    if (nxt[curpos][ch] == 0) {
      nxt[curpos][ch] = lst++;
    }
    curpos = nxt[curpos][ch];
  }
  if (isFin[curpos]) return 0;
  isFin[curpos] = true;
  return 1;
}
int main() {
  for (int i = 0; i < 2005; i++)
    for (int j = 0; j < 2005; j++) good[i][j] = true;
  scanf("%s", s);
  len = strlen(s);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    castRule(i);
  }
  long long ans = 0;
  for (int i = 0; i < len; i++) {
    for (int j = i + 1; j <= len; j++)
      if (good[i][j]) {
        ans += addStr(i, j);
      }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char s[100500];
char c[100500];
char st[100500];
int pf[100500];
bool isEnter[50500];
int len;
int lft[11][50500];
int rght[11][50500];
void calcPf() {
  int len = strlen(c);
  int sWin = 0, eWin = 0;
  for (int i = 1; i < len; i++) {
    int curpf = 0;
    if (eWin > i) {
      curpf = pf[i - sWin];
    }
    curpf = min(curpf, eWin - i);
    curpf = max(0, curpf);
    while (c[i + curpf] == c[curpf]) curpf++;
    if (i + curpf > eWin) {
      eWin = i + curpf;
      sWin = i;
    }
    pf[i] = curpf;
  }
}
int cnt[50500];
bool good[2005][2005];
void castRule(int rule) {
  memset(isEnter, false, sizeof isEnter);
  int a, b;
  scanf("%s %d %d", st, &a, &b);
  int curLen = strlen(st);
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) c[j - i] = s[j];
    c[len - i] = '$';
    for (int j = 0; j < curLen; j++) c[j + len - i + 1] = st[j];
    c[curLen + len - i + 1] = 0;
    calcPf();
    for (int j = 0; j < len; j++) cnt[j] = 0;
    for (int j = 0; j < curLen; j++) cnt[pf[j + len - i + 1]]++;
    int curCnt = curLen;
    for (int j = i; j <= len; j++) {
      if (curCnt < a || curCnt > b) good[i][j] = false;
      curCnt -= cnt[j - i];
    }
  }
}
int nxt[2000500][26];
bool isFin[2000500];
int lst = 1;
int addStr(int l, int r) {
  int curpos = 0;
  for (int i = l; i < r; i++) {
    char ch = s[i] - 'a';
    if (nxt[curpos][ch] == 0) {
      nxt[curpos][ch] = lst++;
    }
    curpos = nxt[curpos][ch];
  }
  if (isFin[curpos]) return 0;
  isFin[curpos] = true;
  return 1;
}
int main() {
  for (int i = 0; i < 2005; i++)
    for (int j = 0; j < 2005; j++) good[i][j] = true;
  scanf("%s", s);
  len = strlen(s);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    castRule(i);
  }
  long long ans = 0;
  for (int i = 0; i < len; i++) {
    for (int j = i + 1; j <= len; j++)
      if (good[i][j]) {
        ans += addStr(i, j);
      }
  }
  printf("%I64d\n", ans);
  return 0;
}

```