#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 1e6 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = 1ll * a * a % M)
    if (b & 1) ret = 1ll * ret * a % M;
  return ret;
}
struct SAM {
  int next[maxn][47], fail[maxn], len[maxn];
  int cnt, last;
  void init() {
    cnt = last = 0;
    fail[0] = -1;
    len[0] = 0;
    memset(next[0], 0, sizeof(next[0]));
  }
  int CNT[maxn];
  bool OK[maxn];
  void add(int c, bool mark) {
    int np = ++cnt, p = last;
    memset(next[np], 0, sizeof(next[np]));
    len[np] = len[p] + 1;
    CNT[np] = mark;
    OK[np] = !mark;
    for (; p != -1 && !next[p][c]; p = fail[p]) next[p][c] = np;
    if (p == -1)
      fail[np] = 0;
    else {
      int q = next[p][c];
      if (len[p] + 1 == len[q])
        fail[np] = q;
      else {
        int nq = ++cnt;
        len[nq] = len[p] + 1;
        memcpy(next[nq], next[q], sizeof(next[q]));
        fail[nq] = fail[q];
        fail[np] = fail[q] = nq;
        for (; p != -1 && next[p][c] == q; p = fail[p]) next[p][c] = nq;
        OK[nq] = OK[q];
        CNT[nq] = 0;
      }
    }
    last = np;
  }
  int S[maxn], Q[maxn];
  void solve(int l, int r) {
    int i;
    for (i = 1; i <= cnt; i++) S[i] = 0;
    for (i = 1; i <= cnt; i++) S[len[i]]++;
    for (i = 1; i <= cnt; i++) S[i] += S[i - 1];
    for (i = 1; i <= cnt; i++) Q[S[len[i]]--] = i;
    for (i = cnt; i >= 1; i--) CNT[fail[Q[i]]] += CNT[Q[i]];
    for (i = 1; i <= cnt; i++)
      if (CNT[i] < l || r < CNT[i]) OK[i] = 0;
    for (i = 1; i <= cnt; i++) CNT[i] = 0;
  }
  long long Solve() {
    int i;
    long long ret = 0;
    for (i = 1; i <= cnt; i++)
      if (OK[i]) ret += len[i] - len[fail[i]];
    return ret;
  }
  char a[maxn];
  void dfs(int x = 0, int len = 0) {
    int i;
    printf("%-3d %-3d ", CNT[x], OK[x]);
    printf("%-3d(fail:%-3d):%s\n", x, fail[x], a);
    for (i = 0; i < 26; i++) {
      if (next[x][i]) {
        a[len] = i + 'a';
        dfs(next[x][i], len + 1);
        a[len] = 0;
      }
    }
  }
} sam;
int n, m, T;
int i, j, k;
char a[maxn];
int main() {
  scanf("%s", a);
  n = strlen(a);
  sam.init();
  for (i = 0; i < n; i++) sam.add(a[i] - 'a', 0);
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    sam.add(26 + i, 1);
    int l, r;
    scanf("%s%d%d", a, &l, &r);
    int i, n = strlen(a);
    for (i = 0; i < n; i++) sam.add(a[i] - 'a', 1);
    sam.solve(l, r);
  }
  printf("%I64d\n", sam.Solve());
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 1e6 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = 1ll * a * a % M)
    if (b & 1) ret = 1ll * ret * a % M;
  return ret;
}
struct SAM {
  int next[maxn][47], fail[maxn], len[maxn];
  int cnt, last;
  void init() {
    cnt = last = 0;
    fail[0] = -1;
    len[0] = 0;
    memset(next[0], 0, sizeof(next[0]));
  }
  int CNT[maxn];
  bool OK[maxn];
  void add(int c, bool mark) {
    int np = ++cnt, p = last;
    memset(next[np], 0, sizeof(next[np]));
    len[np] = len[p] + 1;
    CNT[np] = mark;
    OK[np] = !mark;
    for (; p != -1 && !next[p][c]; p = fail[p]) next[p][c] = np;
    if (p == -1)
      fail[np] = 0;
    else {
      int q = next[p][c];
      if (len[p] + 1 == len[q])
        fail[np] = q;
      else {
        int nq = ++cnt;
        len[nq] = len[p] + 1;
        memcpy(next[nq], next[q], sizeof(next[q]));
        fail[nq] = fail[q];
        fail[np] = fail[q] = nq;
        for (; p != -1 && next[p][c] == q; p = fail[p]) next[p][c] = nq;
        OK[nq] = OK[q];
        CNT[nq] = 0;
      }
    }
    last = np;
  }
  int S[maxn], Q[maxn];
  void solve(int l, int r) {
    int i;
    for (i = 1; i <= cnt; i++) S[i] = 0;
    for (i = 1; i <= cnt; i++) S[len[i]]++;
    for (i = 1; i <= cnt; i++) S[i] += S[i - 1];
    for (i = 1; i <= cnt; i++) Q[S[len[i]]--] = i;
    for (i = cnt; i >= 1; i--) CNT[fail[Q[i]]] += CNT[Q[i]];
    for (i = 1; i <= cnt; i++)
      if (CNT[i] < l || r < CNT[i]) OK[i] = 0;
    for (i = 1; i <= cnt; i++) CNT[i] = 0;
  }
  long long Solve() {
    int i;
    long long ret = 0;
    for (i = 1; i <= cnt; i++)
      if (OK[i]) ret += len[i] - len[fail[i]];
    return ret;
  }
  char a[maxn];
  void dfs(int x = 0, int len = 0) {
    int i;
    printf("%-3d %-3d ", CNT[x], OK[x]);
    printf("%-3d(fail:%-3d):%s\n", x, fail[x], a);
    for (i = 0; i < 26; i++) {
      if (next[x][i]) {
        a[len] = i + 'a';
        dfs(next[x][i], len + 1);
        a[len] = 0;
      }
    }
  }
} sam;
int n, m, T;
int i, j, k;
char a[maxn];
int main() {
  scanf("%s", a);
  n = strlen(a);
  sam.init();
  for (i = 0; i < n; i++) sam.add(a[i] - 'a', 0);
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    sam.add(26 + i, 1);
    int l, r;
    scanf("%s%d%d", a, &l, &r);
    int i, n = strlen(a);
    for (i = 0; i < n; i++) sam.add(a[i] - 'a', 1);
    sam.solve(l, r);
  }
  printf("%I64d\n", sam.Solve());
}

```