We call a positive integer x a k-beautiful integer if and only if it is possible to split the multiset of its digits in the decimal representation into two subsets such that the difference between the sum of digits in one subset and the sum of digits in the other subset is less than or equal to k. Each digit should belong to exactly one subset after the split.

There are n queries for you. Each query is described with three integers l, r and k, which mean that you are asked how many integers x between l and r (inclusive) are k-beautiful.

Input

The first line contains a single integer n (1 ≤ n ≤ 5·104), indicating the number of queries.

Each of the next n lines describes a query, containing three integers l, r and k (1 ≤ l ≤ r ≤ 1018, 0 ≤ k ≤ 9).

Output

For each query print a single number — the answer to the query.

Examples

Input

10
1 100 0
1 100 1
1 100 2
1 100 3
1 100 4
1 100 5
1 100 6
1 100 7
1 100 8
1 100 9


Output

9
28
44
58
70
80
88
94
98
100


Input

10
1 1000 0
1 1000 1
1 1000 2
1 1000 3
1 1000 4
1 1000 5
1 1000 6
1 1000 7
1 1000 8
1 1000 9


Output

135
380
573
721
830
906
955
983
996
1000

Note

If 1 ≤ x ≤ 9, integer x is k-beautiful if and only if x ≤ k.

If 10 ≤ x ≤ 99, integer x = 10a + b is k-beautiful if and only if |a - b| ≤ k, where a and b are integers between 0 and 9, inclusive.

100 is k-beautiful if and only if k ≥ 1.
#include <bits/stdc++.h>
using namespace std;
void upd(long long &x, long long y) { x += y; }
struct node {
  int nxt;
  long long sta, v[10];
  node() {
    nxt = 0;
    memset(v, 0, sizeof(v));
  }
};
struct state {
  vector<node> nodes;
  int head[100007];
  state() { memset(head, -1, sizeof(head)); }
  size_t size() const { return nodes.size(); }
  void sum() {
    for (int i = 0; i < nodes.size(); i++)
      for (int j = 1; j <= 9; j++) upd(nodes[i].v[j], nodes[i].v[j - 1]);
  }
  void update(int at, long long key, long long v) {
    int tmp = key % 100007;
    for (int i = head[tmp]; i != -1; i = nodes[i].nxt)
      if (nodes[i].sta == key) {
        upd(nodes[i].v[at], v);
        return;
      }
    node now;
    now.nxt = head[tmp];
    head[tmp] = nodes.size();
    now.v[at] = v;
    now.sta = key;
    nodes.push_back(now);
  }
  long long find(int at, long long key) {
    int tmp = key % 100007;
    for (int i = head[tmp]; i != -1; i = nodes[i].nxt)
      if (nodes[i].sta == key) return nodes[i].v[at];
    return 0;
  }
} f[11][20];
long long c[2][20], bit[15];
bitset<200> vis[15], pre[200], tmp;
struct big {
  int a[35], len;
  void read() {
    memset(a, 0, sizeof(a));
    len = 0;
    char ch = getchar();
    for (; ch < '0' || ch > '9'; ch = getchar())
      ;
    for (; ch >= '0' && ch <= '9'; ch = getchar()) a[len++] = ch - 48;
  }
  long long mod() {
    long long ans = 0;
    for (int i = 0; i < len; i++)
      ans = (10ll * ans + a[i]) % 1000000000000000007ll;
    return ans;
  }
  void plus() {
    a[len - 1]++;
    for (int i = len - 1; i >= 1; i--) a[i - 1] += a[i] / 10, a[i] %= 10;
    if (a[0] > 9) {
      a[0] = 1;
      for (int i = 1; i <= len; i++) a[i] = 0;
      len++;
    }
  }
} L, R;
int sum[40], times = 0;
void dfs(int dep, int cnt, int sum, long long msk) {
  int mx = sum + (18 - cnt) * 9;
  tmp = pre[mx >> 1];
  if (sum >> 1) tmp ^= pre[(sum >> 1) - 1];
  if ((vis[dep - 1] & tmp) == tmp) return;
  if (dep == 10) {
    for (int i = sum >> 1; i >= 0; i--)
      if (vis[dep - 1][i]) {
        int dif = sum - i - i;
        if (dif > 1) {
          ::sum[cnt]++;
          f[dif][0].update(0, msk, 1);
        }
        break;
      }
    return;
  }
  vis[dep] = vis[dep - 1];
  dfs(dep + 1, cnt, sum, msk);
  for (; cnt < 18;) {
    vis[dep] |= vis[dep] << dep;
    dfs(dep + 1, ++cnt, sum += dep, msk += bit[dep]);
  }
}
long long solve(int dif, big lim) {
  long long ans = 0, msk = 0;
  int len = lim.len;
  state *F = f[dif];
  for (int i = 0; i < len; i++)
    if (len - i > 18) {
      for (int j = 0; j < lim.a[i]; j++) ans += F[len - i - 1].find(9, msk);
    } else if (lim.a[i]) {
      ans += F[len - i].find(lim.a[i] - 1, msk);
      msk += bit[lim.a[i]];
    }
  return ans;
}
long long solve2(big lim) {
  int len = lim.len;
  long long ans = 0;
  int part = 0;
  for (int i = 0; i < len; i++) {
    int odd = lim.a[i] >> 1, even = lim.a[i] - odd;
    ans += c[part][len - 1 - i] * odd + c[part ^ 1][len - 1 - i] * even;
    part ^= lim.a[i] & 1;
  }
  return ans;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 - 48 + ch;
  return x;
}
int main() {
  pre[0][0] = 1;
  for (int i = 1; i <= 190; i++) pre[i] = pre[i - 1], pre[i][i] = 1;
  c[0][0] = 1;
  for (int i = 1; i <= 18; i++)
    c[0][i] = c[1][i] =
        5ll * (c[0][i - 1] + c[1][i - 1]) % 1000000000000000007ll;
  bit[1] = 1;
  for (int i = 2; i < 10; i++) bit[i] = bit[i - 1] << 5;
  vis[0].set(0);
  dfs(1, 0, 0, 0);
  for (int i = 2; i < 10; i++) {
    f[i][0].sum();
    for (int j = 0; j < 18; j++) {
      state &cur = f[i][j], &nxt = f[i][j + 1];
      for (int id = 0, sz = cur.size(); id < sz; id++) {
        int cnt = j;
        long long way = cur.nodes[id].v[9];
        long long msk = cur.nodes[id].sta, tmp = msk;
        for (int k = 1; k < 10; k++, tmp >>= 5) {
          int rem = tmp & 31;
          if (!rem) continue;
          cnt += rem;
          nxt.update(k, msk - bit[k], way);
        }
        if (cnt < 18) nxt.update(0, msk, way);
      }
      nxt.sum();
    }
  }
  int Q = read();
  while (Q--) {
    L.read();
    R.read();
    R.plus();
    int k = read();
    long long ans = R.mod() - L.mod();
    if (!k) {
      ans -= solve2(R) - solve2(L);
      for (int i = 2; i < 10; i += 2) ans -= solve(i, R) - solve(i, L);
    } else
      for (int i = k + 1; i < 10; i++) ans -= solve(i, R) - solve(i, L);
    printf("%lld\n", (ans % 1000000000000000007ll + 1000000000000000007ll) %
                         1000000000000000007ll);
  }
  return 0;
}
