#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const double eps = 1e-8;
const int inf = 1e9 + 7;
using namespace std;
struct edge {
  int t;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y) {
  o->t = y;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
char str[MAXN];
char ge[11];
int rt, cur, cnt;
int L[12], R[12];
int pre[MAXN], dis[MAXN], fa[MAXN], sz[MAXN][11], ch[MAXN][36];
int ans[MAXN], pos[MAXN], maxx[MAXN];
bool flag[MAXN];
void built(int x, int id) {
  int last = cur;
  cur = ++cnt;
  dis[cur] = dis[last] + 1;
  int p = last;
  sz[cur][id] = 1;
  for (; p && !ch[p][x]; p = fa[p]) ch[p][x] = cur;
  if (!p)
    fa[cur] = rt;
  else {
    int q = ch[p][x];
    if (dis[q] == dis[p] + 1)
      fa[cur] = q;
    else {
      int nt = ++cnt;
      dis[nt] = dis[p] + 1;
      memcpy(ch[nt], ch[q], sizeof(ch[q]));
      fa[nt] = fa[q];
      fa[q] = fa[cur] = nt;
      for (; ch[p][x] == q; p = fa[p]) ch[p][x] = nt;
    }
  }
}
void dfs(int x) {
  for (edge *j = h[x]; j; j = j->next) {
    dfs(j->t);
    if (!flag[x] && flag[j->t]) flag[x] = 1;
    for (int i = 1; i <= 10; i++) sz[x][i] += sz[j->t][i];
  }
}
char s2[MAXN];
int main() {
  scanf("%s", str);
  int n = read();
  int len = strlen(str);
  rt = cnt = cur = 1;
  for (int i = 0; i <= len - 1; i++) built(str[i] - 'a', 0);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d%d", s2, &L[i], &R[i]);
    int len1 = strlen(s2);
    cur = 1;
    for (int j = 0; j <= len1 - 1; j++) built(s2[j] - 'a', i);
  }
  if (!n) {
    long long ans1 = 0;
    for (int i = 2; i <= cnt; i++) ans1 += dis[i] - dis[fa[i]];
    printf("%lld\n", ans1);
    return 0;
  }
  int temp = rt;
  for (int i = 0; i < len; i++) {
    int t1 = str[i] - 'a';
    temp = ch[temp][t1];
    flag[temp] = 1;
  }
  for (int i = 1; i <= cnt; i++) add(fa[i], i);
  dfs(rt);
  long long ans1 = 0;
  for (int i = 2; i <= cnt; i++) {
    if (!flag[i]) continue;
    bool flag1 = 0;
    for (int j = 1; j <= n; j++) {
      if (sz[i][j] < L[j] || sz[i][j] > R[j]) {
        flag1 = 1;
        break;
      }
    }
    if (!flag1) ans1 += dis[i] - dis[fa[i]];
  }
  printf("%lld\n", ans1);
  return 0;
}

### Prompt

Generate a Cpp solution 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>
const int MAXN = 1e5 + 10;
const double eps = 1e-8;
const int inf = 1e9 + 7;
using namespace std;
struct edge {
  int t;
  edge *next;
} e[MAXN << 1], *h[MAXN], *o = e;
void add(int x, int y) {
  o->t = y;
  o->next = h[x];
  h[x] = o++;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
char str[MAXN];
char ge[11];
int rt, cur, cnt;
int L[12], R[12];
int pre[MAXN], dis[MAXN], fa[MAXN], sz[MAXN][11], ch[MAXN][36];
int ans[MAXN], pos[MAXN], maxx[MAXN];
bool flag[MAXN];
void built(int x, int id) {
  int last = cur;
  cur = ++cnt;
  dis[cur] = dis[last] + 1;
  int p = last;
  sz[cur][id] = 1;
  for (; p && !ch[p][x]; p = fa[p]) ch[p][x] = cur;
  if (!p)
    fa[cur] = rt;
  else {
    int q = ch[p][x];
    if (dis[q] == dis[p] + 1)
      fa[cur] = q;
    else {
      int nt = ++cnt;
      dis[nt] = dis[p] + 1;
      memcpy(ch[nt], ch[q], sizeof(ch[q]));
      fa[nt] = fa[q];
      fa[q] = fa[cur] = nt;
      for (; ch[p][x] == q; p = fa[p]) ch[p][x] = nt;
    }
  }
}
void dfs(int x) {
  for (edge *j = h[x]; j; j = j->next) {
    dfs(j->t);
    if (!flag[x] && flag[j->t]) flag[x] = 1;
    for (int i = 1; i <= 10; i++) sz[x][i] += sz[j->t][i];
  }
}
char s2[MAXN];
int main() {
  scanf("%s", str);
  int n = read();
  int len = strlen(str);
  rt = cnt = cur = 1;
  for (int i = 0; i <= len - 1; i++) built(str[i] - 'a', 0);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d%d", s2, &L[i], &R[i]);
    int len1 = strlen(s2);
    cur = 1;
    for (int j = 0; j <= len1 - 1; j++) built(s2[j] - 'a', i);
  }
  if (!n) {
    long long ans1 = 0;
    for (int i = 2; i <= cnt; i++) ans1 += dis[i] - dis[fa[i]];
    printf("%lld\n", ans1);
    return 0;
  }
  int temp = rt;
  for (int i = 0; i < len; i++) {
    int t1 = str[i] - 'a';
    temp = ch[temp][t1];
    flag[temp] = 1;
  }
  for (int i = 1; i <= cnt; i++) add(fa[i], i);
  dfs(rt);
  long long ans1 = 0;
  for (int i = 2; i <= cnt; i++) {
    if (!flag[i]) continue;
    bool flag1 = 0;
    for (int j = 1; j <= n; j++) {
      if (sz[i][j] < L[j] || sz[i][j] > R[j]) {
        flag1 = 1;
        break;
      }
    }
    if (!flag1) ans1 += dis[i] - dis[fa[i]];
  }
  printf("%lld\n", ans1);
  return 0;
}

```