#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC target("avx")
#pragma GCC optimize("Ofast")
using namespace std;
const unsigned _mod = 998244353;
const unsigned mod = 1e9 + 7;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
long long qpow(long long x, long long y, long long m) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % m;
    y >>= 1ll;
    x = x * x % m;
  }
  return ret;
}
long long _gcd(long long x, long long y) { return y ? _gcd(y, x % y) : x; }
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
char s[1000010];
int q, h, t, n, m = 1, cur = 1, ans, l[33], r[33], c[500010][33], a[500010][33],
                g[500010], fa[500010], d[500010], f[500010];
void jyz(long long i, long long t, long long e) {
  long long x = cur, y = cur = ++m;
  g[y] = i, ++c[y][t];
  while (x && !a[x][e]) a[x][e] = y, x = fa[x];
  if (!x)
    fa[y] = 1;
  else {
    long long z = a[x][e];
    if (g[z] == g[x] + 1)
      fa[y] = z;
    else {
      long long u = ++m;
      g[u] = g[x] + 1;
      memcpy(a[u], a[z], sizeof(a[u]));
      fa[u] = fa[z], fa[z] = fa[y] = u;
      while (x && a[x][e] == z) a[x][e] = u, x = fa[x];
    }
  }
}
bool txdy(long long x) {
  if (!*c[x]) return 0;
  for (register int i = 1; i <= n; ++i)
    if (l[i] > c[x][i] || c[x][i] > r[i]) return 0;
  return 1;
}
int main() {
  scanf("%s", s + 1);
  cin >> n;
  q = strlen(s + 1);
  for (register int i = 1; i <= n; ++i)
    scanf("%s", s + q + 2), l[i] = read(), r[i] = read(),
                            q += strlen(s + q + 2) + 1;
  for (register int i = 1; i <= q; ++i) {
    if (!s[i])
      jyz(i, ++t, 0);
    else
      jyz(i, t, s[i] - 96);
  }
  for (register int i = 1; i <= m; ++i) ++d[fa[i]];
  h = 0, t = 0;
  for (register int i = 1; i <= m; ++i)
    if (!d[i]) f[++t] = i;
  while (h < t) {
    long long x = f[++h], y = fa[x];
    ans += (g[x] - g[y]) * txdy(x);
    *c[y] |= *c[x];
    for (register int i = 1; i <= n; ++i) c[y][i] += c[x][i];
    if (!--d[y]) f[++t] = y;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize(3)
#pragma GCC target("avx")
#pragma GCC optimize("Ofast")
using namespace std;
const unsigned _mod = 998244353;
const unsigned mod = 1e9 + 7;
const long long infi = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
long long ksm(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % mod;
    y >>= 1ll;
    x = x * x % mod;
  }
  return ret;
}
long long qpow(long long x, long long y, long long m) {
  long long ret = 1;
  while (y) {
    if (y & 1ll) ret = ret * x % m;
    y >>= 1ll;
    x = x * x % m;
  }
  return ret;
}
long long _gcd(long long x, long long y) { return y ? _gcd(y, x % y) : x; }
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
char s[1000010];
int q, h, t, n, m = 1, cur = 1, ans, l[33], r[33], c[500010][33], a[500010][33],
                g[500010], fa[500010], d[500010], f[500010];
void jyz(long long i, long long t, long long e) {
  long long x = cur, y = cur = ++m;
  g[y] = i, ++c[y][t];
  while (x && !a[x][e]) a[x][e] = y, x = fa[x];
  if (!x)
    fa[y] = 1;
  else {
    long long z = a[x][e];
    if (g[z] == g[x] + 1)
      fa[y] = z;
    else {
      long long u = ++m;
      g[u] = g[x] + 1;
      memcpy(a[u], a[z], sizeof(a[u]));
      fa[u] = fa[z], fa[z] = fa[y] = u;
      while (x && a[x][e] == z) a[x][e] = u, x = fa[x];
    }
  }
}
bool txdy(long long x) {
  if (!*c[x]) return 0;
  for (register int i = 1; i <= n; ++i)
    if (l[i] > c[x][i] || c[x][i] > r[i]) return 0;
  return 1;
}
int main() {
  scanf("%s", s + 1);
  cin >> n;
  q = strlen(s + 1);
  for (register int i = 1; i <= n; ++i)
    scanf("%s", s + q + 2), l[i] = read(), r[i] = read(),
                            q += strlen(s + q + 2) + 1;
  for (register int i = 1; i <= q; ++i) {
    if (!s[i])
      jyz(i, ++t, 0);
    else
      jyz(i, t, s[i] - 96);
  }
  for (register int i = 1; i <= m; ++i) ++d[fa[i]];
  h = 0, t = 0;
  for (register int i = 1; i <= m; ++i)
    if (!d[i]) f[++t] = i;
  while (h < t) {
    long long x = f[++h], y = fa[x];
    ans += (g[x] - g[y]) * txdy(x);
    *c[y] |= *c[x];
    for (register int i = 1; i <= n; ++i) c[y][i] += c[x][i];
    if (!--d[y]) f[++t] = y;
  }
  cout << ans << '\n';
  return 0;
}

```