#include <bits/stdc++.h>
using namespace std;
long long a[1010000], b[1010000], tms[1010000], C[1010000];
const long long mod = 1000000007;
long long qpow(long long x, long long y, const long long& k) {
  long long t = 1;
  while (y) {
    if (y & 1) {
      t = t * x % k;
    }
    x = x * x % k;
    y >>= 1;
  }
  return t;
}
long long inv(long long x) { return qpow(x, mod - 2, mod); }
signed main() {
  std::ios::sync_with_stdio(false);
  long long n, l, k;
  cin >> n >> l >> k;
  long long tk = l / n;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  long long h = 0;
  sort(b + 1, b + n + 1);
  for (long long i = 1; i <= n; ++i) {
    if (b[i] != b[i - 1]) b[++h] = b[i];
    ++tms[h];
  }
  for (long long i = 1; i <= n; ++i)
    a[i] = lower_bound(b + 1, b + h + 1, a[i]) - b;
  vector<vector<long long>> f(k + 5);
  for (auto& i : f) i = vector<long long>(h + 5);
  vector<vector<long long>> lst(k + 5);
  for (auto& i : lst) i = vector<long long>(h + 5);
  C[0] = 1;
  for (long long j = 0; j <= h; ++j) lst[0][j] = 1;
  for (long long i = 1; i <= min(k, tk); ++i) {
    for (long long j = 1; j <= h; ++j) {
      f[i][j] = lst[i - 1][j] * tms[j] % mod;
      lst[i][j] = (lst[i][j - 1] + f[i][j]) % mod;
      (f[i][j] *= (tk - i + 1) % mod) %= mod;
    }
  }
  long long rem = l % n;
  if (rem) {
    f[k + 1] = vector<long long>(rem + 5);
    for (long long j = 1; j <= rem; ++j) {
      for (long long r = 0; r < k; ++r) (f[k + 1][j] += lst[r][a[j]]) %= mod;
    }
  }
  long long ans = 0;
  for (auto& p : f) {
    for (auto& q : p) {
      ans = (ans + q) % mod;
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long a[1010000], b[1010000], tms[1010000], C[1010000];
const long long mod = 1000000007;
long long qpow(long long x, long long y, const long long& k) {
  long long t = 1;
  while (y) {
    if (y & 1) {
      t = t * x % k;
    }
    x = x * x % k;
    y >>= 1;
  }
  return t;
}
long long inv(long long x) { return qpow(x, mod - 2, mod); }
signed main() {
  std::ios::sync_with_stdio(false);
  long long n, l, k;
  cin >> n >> l >> k;
  long long tk = l / n;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  long long h = 0;
  sort(b + 1, b + n + 1);
  for (long long i = 1; i <= n; ++i) {
    if (b[i] != b[i - 1]) b[++h] = b[i];
    ++tms[h];
  }
  for (long long i = 1; i <= n; ++i)
    a[i] = lower_bound(b + 1, b + h + 1, a[i]) - b;
  vector<vector<long long>> f(k + 5);
  for (auto& i : f) i = vector<long long>(h + 5);
  vector<vector<long long>> lst(k + 5);
  for (auto& i : lst) i = vector<long long>(h + 5);
  C[0] = 1;
  for (long long j = 0; j <= h; ++j) lst[0][j] = 1;
  for (long long i = 1; i <= min(k, tk); ++i) {
    for (long long j = 1; j <= h; ++j) {
      f[i][j] = lst[i - 1][j] * tms[j] % mod;
      lst[i][j] = (lst[i][j - 1] + f[i][j]) % mod;
      (f[i][j] *= (tk - i + 1) % mod) %= mod;
    }
  }
  long long rem = l % n;
  if (rem) {
    f[k + 1] = vector<long long>(rem + 5);
    for (long long j = 1; j <= rem; ++j) {
      for (long long r = 0; r < k; ++r) (f[k + 1][j] += lst[r][a[j]]) %= mod;
    }
  }
  long long ans = 0;
  for (auto& p : f) {
    for (auto& q : p) {
      ans = (ans + q) % mod;
    }
  }
  cout << ans << '\n';
  return 0;
}

```