Problem: What-The-Fatherland is a strange country! All phone numbers there are strings consisting of lowercase English letters. What is double strange that a phone number can be associated with several bears!

In that country there is a rock band called CF consisting of n bears (including Mike) numbered from 1 to n. 

<image>

Phone number of i-th member of CF is si. May 17th is a holiday named Phone Calls day. In the last Phone Calls day, everyone called all the numbers that are substrings of his/her number (one may call some number several times). In particular, everyone called himself (that was really strange country).

Denote as call(i, j) the number of times that i-th member of CF called the j-th member of CF. 

The geek Mike has q questions that he wants to ask you. In each question he gives you numbers l, r and k and you should tell him the number 

<image>

Input

The first line of input contains integers n and q (1 ≤ n ≤ 2 × 105 and 1 ≤ q ≤ 5 × 105).

The next n lines contain the phone numbers, i-th line contains a string si consisting of lowercase English letters (<image>).

The next q lines contain the information about the questions, each of them contains integers l, r and k (1 ≤ l ≤ r ≤ n and 1 ≤ k ≤ n).

Output

Print the answer for each question in a separate line.

Examples

Input

5 5
a
ab
abab
ababab
b
1 5 1
3 5 1
1 5 2
1 5 3
1 4 5


Output

7
5
6
3
6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T alex(T x) {
  return (x > 0) ? x : -x;
}
template <class T>
void read(T& x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while ((ch < 48 || ch > 57) && ch != 45) ch = getchar();
  if (ch == 45) f = -1, ch = getchar();
  while (ch >= 48 && ch <= 57) x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
int n, q;
char s[200005];
int t[200005 << 1];
int st[200005], ed[200005], len;
int tmp[200005 << 1], sa[200005 << 1], rnk[200005 << 1], sum[200005 << 1];
bool cmp(int* tmp, int x, int y, int l) {
  return tmp[x] == tmp[y] && tmp[x + l] == tmp[y + l];
}
void getsa(int n, int m) {
  memset(sum, 0, sizeof(sum));
  for (int i = (1); i <= (n); i++) sum[rnk[i] = t[i]]++;
  for (int i = (1); i <= (m); i++) sum[i] += sum[i - 1];
  for (int i = (n); i >= (1); i--) sa[sum[rnk[i]]--] = i;
  for (int j = 1; j <= n; j <<= 1) {
    int p = 0;
    for (int i = (n); i >= (n - j + 1); i--) tmp[++p] = i;
    for (int i = (1); i <= (n); i++)
      if (sa[i] > j) tmp[++p] = sa[i] - j;
    memset(sum, 0, sizeof(sum));
    for (int i = (1); i <= (n); i++) sum[rnk[tmp[i]]]++;
    for (int i = (1); i <= (max(n, m)); i++) sum[i] += sum[i - 1];
    for (int i = (n); i >= (1); i--) sa[sum[rnk[tmp[i]]]--] = tmp[i];
    for (int i = (1); i <= (n); i++) tmp[i] = rnk[i];
    p = 1;
    rnk[sa[1]] = 1;
    for (int i = (2); i <= (n); i++)
      rnk[sa[i]] = p += !cmp(tmp, sa[i - 1], sa[i], j);
  }
}
int height[200005 << 1][20];
void getheight(int n) {
  for (int i = (1); i <= (n); i++) {
    int l = max(height[rnk[i - 1]][0] - 1, 0), j = sa[rnk[i] - 1];
    while (i + l <= n && j + l <= n && t[i + l] == t[j + l]) l++;
    height[rnk[i]][0] = l;
  }
  for (int j = (1); j < (20); j++)
    for (int i = (1); i <= (n - (1 << j) + 1); i++)
      height[i][j] = min(height[i][j - 1], height[i + (1 << (j - 1))][j - 1]);
}
int _log2[200005 << 1];
int LCP(int x, int y) {
  if (x == y) return n - sa[x] + 1;
  if (x > y) swap(x, y);
  x++;
  int j = _log2[y - x + 1];
  return min(height[x][j], height[y - (1 << j) + 1][j]);
}
struct Node {
  int idx, l, sign;
  Node() {}
  Node(int idx, int l, int sign) : idx(idx), l(l), sign(sign) {}
};
int sum_bit[200005 << 1];
int lowbit(int x) { return x & (-x); }
void insert(int v, int x) {
  for (int i = x; i <= len; i += lowbit(i)) sum_bit[i]++;
}
int ask(int l, int r) {
  int res = 0;
  for (int i = l - 1; i; i -= lowbit(i)) res -= sum_bit[i];
  for (int i = r; i; i -= lowbit(i)) res += sum_bit[i];
  return res;
}
vector<Node> que[200005];
int ans[500005];
void insert(int idx) {
  for (int i = (st[idx]); i <= (ed[idx]); i++) insert(1, rnk[i]);
}
int findl(int l, int r, int pos, int val) {
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (LCP(mid, pos) >= val)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return l;
}
int findr(int l, int r, int pos, int val) {
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (LCP(pos, mid) >= val)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int query(int idx) {
  int len2 = ed[idx] - st[idx] + 1;
  int l = findl(1, rnk[st[idx]] - 1, rnk[st[idx]], len2);
  int r = findr(rnk[st[idx]] + 1, len, rnk[st[idx]], len2);
  return ask(l, r);
}
int main() {
  read(n);
  read(q);
  int m = 26;
  len = 0;
  for (int i = (1); i <= (n); i++) {
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    st[i] = len + 1;
    for (int k = (1); k <= (l); k++) t[++len] = s[k] - 'a' + 1;
    ed[i] = len;
    t[++len] = ++m;
  }
  getsa(len, m);
  getheight(len);
  for (int i = (0); i <= (len); i++) _log2[i] = log2(i);
  for (int i = (1); i <= (q); i++) {
    int l, r, k;
    read(l);
    read(r);
    read(k);
    que[l - 1].push_back(Node(i, k, -1));
    que[r].push_back(Node(i, k, 1));
  }
  for (int i = (1); i <= (n); i++) {
    insert(i);
    for (int k = (0); k < (que[i].size()); k++)
      if (que[i][k].sign == -1)
        ans[que[i][k].idx] -= query(que[i][k].l);
      else
        ans[que[i][k].idx] += query(que[i][k].l);
  }
  for (int i = (1); i <= (q); i++) printf("%d\n", ans[i]);
  return 0;
}