#include <bits/stdc++.h>
using namespace std;
struct E {
  int v, n;
} G[1000010];
int m, k;
int cnt, point[1000010];
int pl[12], pr[12];
void adde(int u, int v) { G[++cnt] = (E){v, point[u]}, point[u] = cnt; }
long long ans;
bool vis[1000010];
struct SAM {
  static const int K = 26;
  int rt, la, nodes;
  int len[1000010], n[1000010][K + 1], pa[1000010], sz[12][1000010];
  void init() {
    nodes = 0;
    rt = la = newnode(0);
  }
  int newnode(int pl) {
    int i = ++nodes;
    len[i] = pl;
    return i;
  }
  void ins(int w, int d) {
    int p = la, np = newnode(len[p] + 1);
    la = np;
    if (w < 26) sz[d][np] = 1;
    while (p && !n[p][w]) n[p][w] = np, p = pa[p];
    if (!p)
      pa[np] = rt;
    else {
      int q = n[p][w];
      if (len[q] == len[p] + 1)
        pa[np] = q;
      else {
        int nq = newnode(len[p] + 1);
        memcpy(n[nq], n[q], sizeof(n[q]));
        pa[nq] = pa[q];
        pa[q] = pa[np] = nq;
        while (p && n[p][w] == q) n[p][w] = nq, p = pa[p];
      }
    }
  }
  void getsz() {
    for (int i = 2, lim = nodes; i <= lim; ++i) adde(pa[i], i);
    dfs(rt);
  }
  void dfs1(int u) {
    vis[u] = 1;
    for (int i = 0, lim = K - 1; i <= lim; ++i) {
      if (n[u][i]) dfs1(n[u][i]);
    }
  }
  void dfs(int u) {
    for (int i = point[u]; i; i = G[i].n) {
      int v = G[i].v;
      dfs(v);
      for (int j = 0, lim = k; j <= lim; ++j) sz[j][u] += sz[j][v];
    }
    if (sz[0][u]) {
      int ok = 1;
      for (int j = 0, lim = k - 1; j <= lim; ++j)
        if (sz[j + 1][u] > pr[j + 1] || sz[j + 1][u] < pl[j + 1]) ok = 0;
      if (ok) ans += (long long)(len[u] - len[pa[u]]);
    }
  }
  void debug() {
    for (int i = 1, lim = nodes; i <= lim; ++i) {
      printf("i = %d, pa = %d, minlen = %d, maxlen = %d\n", i, pa[i],
             len[pa[i]] + 1, len[i]);
    }
    for (int i = 1, lim = nodes; i <= lim; ++i) {
      for (int j = 0, lim = K; j <= lim; ++j)
        if (n[i][j]) printf("%d %d %c\n", i, n[i][j], j + 'a');
    }
  }
} sam;
char s[1000010];
int main() {
  scanf("%s", s);
  m = strlen(s);
  sam.init();
  for (int j = 0, lim = m - 1; j <= lim; ++j) sam.ins(s[j] - 'a', 0);
  sam.ins(26, 0);
  scanf("%d", &k);
  for (int i = 0, lim = k - 1; i <= lim; ++i) {
    scanf("%s", s);
    scanf("%d", &pl[i + 1]), scanf("%d", &pr[i + 1]);
    m = strlen(s);
    for (int j = 0, lim = m - 1; j <= lim; ++j) sam.ins(s[j] - 'a', i + 1);
    sam.ins(26, i + 1);
  }
  sam.getsz();
  printf("%I64d", (ans));
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
struct E {
  int v, n;
} G[1000010];
int m, k;
int cnt, point[1000010];
int pl[12], pr[12];
void adde(int u, int v) { G[++cnt] = (E){v, point[u]}, point[u] = cnt; }
long long ans;
bool vis[1000010];
struct SAM {
  static const int K = 26;
  int rt, la, nodes;
  int len[1000010], n[1000010][K + 1], pa[1000010], sz[12][1000010];
  void init() {
    nodes = 0;
    rt = la = newnode(0);
  }
  int newnode(int pl) {
    int i = ++nodes;
    len[i] = pl;
    return i;
  }
  void ins(int w, int d) {
    int p = la, np = newnode(len[p] + 1);
    la = np;
    if (w < 26) sz[d][np] = 1;
    while (p && !n[p][w]) n[p][w] = np, p = pa[p];
    if (!p)
      pa[np] = rt;
    else {
      int q = n[p][w];
      if (len[q] == len[p] + 1)
        pa[np] = q;
      else {
        int nq = newnode(len[p] + 1);
        memcpy(n[nq], n[q], sizeof(n[q]));
        pa[nq] = pa[q];
        pa[q] = pa[np] = nq;
        while (p && n[p][w] == q) n[p][w] = nq, p = pa[p];
      }
    }
  }
  void getsz() {
    for (int i = 2, lim = nodes; i <= lim; ++i) adde(pa[i], i);
    dfs(rt);
  }
  void dfs1(int u) {
    vis[u] = 1;
    for (int i = 0, lim = K - 1; i <= lim; ++i) {
      if (n[u][i]) dfs1(n[u][i]);
    }
  }
  void dfs(int u) {
    for (int i = point[u]; i; i = G[i].n) {
      int v = G[i].v;
      dfs(v);
      for (int j = 0, lim = k; j <= lim; ++j) sz[j][u] += sz[j][v];
    }
    if (sz[0][u]) {
      int ok = 1;
      for (int j = 0, lim = k - 1; j <= lim; ++j)
        if (sz[j + 1][u] > pr[j + 1] || sz[j + 1][u] < pl[j + 1]) ok = 0;
      if (ok) ans += (long long)(len[u] - len[pa[u]]);
    }
  }
  void debug() {
    for (int i = 1, lim = nodes; i <= lim; ++i) {
      printf("i = %d, pa = %d, minlen = %d, maxlen = %d\n", i, pa[i],
             len[pa[i]] + 1, len[i]);
    }
    for (int i = 1, lim = nodes; i <= lim; ++i) {
      for (int j = 0, lim = K; j <= lim; ++j)
        if (n[i][j]) printf("%d %d %c\n", i, n[i][j], j + 'a');
    }
  }
} sam;
char s[1000010];
int main() {
  scanf("%s", s);
  m = strlen(s);
  sam.init();
  for (int j = 0, lim = m - 1; j <= lim; ++j) sam.ins(s[j] - 'a', 0);
  sam.ins(26, 0);
  scanf("%d", &k);
  for (int i = 0, lim = k - 1; i <= lim; ++i) {
    scanf("%s", s);
    scanf("%d", &pl[i + 1]), scanf("%d", &pr[i + 1]);
    m = strlen(s);
    for (int j = 0, lim = m - 1; j <= lim; ++j) sam.ins(s[j] - 'a', i + 1);
    sam.ins(26, i + 1);
  }
  sam.getsz();
  printf("%I64d", (ans));
  return 0;
}

```