#include <bits/stdc++.h>
using namespace std;
long long n, m, x, y, z, e, ans, l, k, b[1000100], h[1000100], mod = 1e9 + 7,
                                                               dp[1000100];
pair<long long, long long> a[1000100];
vector<long long> v[1000100];
map<long long, long long> mp;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
    b[i] = a[i].first;
  }
  sort(b, b + n);
  unique(b, b + n);
  for (int i = 0; i < n; i++) {
    mp[b[i]] = i;
    if (b[i] >= b[i + 1] || i == n - 1) {
      m = i + 1;
      break;
    }
  }
  fill(b, b + n, 0);
  for (int i = 0; i < n; i++) {
    a[i].first = mp[a[i].first];
    b[a[i].first]++;
  }
  for (long long i = 0; i < m; i++) {
    y = 1;
    for (long long j = 1; j <= k; j++) {
      (((y += (dp[j - 1]) % mod) %= mod) *= b[i]) %= mod;
      v[i].push_back(y % mod);
    }
    for (long long j = 1; j <= k; j++) {
      (dp[j] += (v[i][j - 1] % mod)) %= mod;
    }
  }
  e = l / n;
  for (long long i = 1; i <= k; i++) {
    if (e < i) break;
    ans += ((dp[i] % mod) *
            (((e - i + 1) % mod) + ((e - i + 1) % mod < 0) * mod)) %
           mod;
    ans %= mod;
  }
  if (l % n != 0) {
    for (long long i = 0; i < n; i++) {
      if (a[i].second < l % n) {
        h[a[i].first]++;
      }
    }
    fill(dp, dp + k + 10, 0);
    for (long long i = 0; i < m; i++) {
      dp[0] = 1;
      for (long long j = 1; j <= k; j++) {
        (dp[j] += (v[i][j - 1] % mod)) %= mod;
      }
      for (long long j = 1; j <= k; j++) {
        if (e >= j - 1) (ans += dp[j - 1] * h[i]) %= mod;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n, m, x, y, z, e, ans, l, k, b[1000100], h[1000100], mod = 1e9 + 7,
                                                               dp[1000100];
pair<long long, long long> a[1000100];
vector<long long> v[1000100];
map<long long, long long> mp;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
    b[i] = a[i].first;
  }
  sort(b, b + n);
  unique(b, b + n);
  for (int i = 0; i < n; i++) {
    mp[b[i]] = i;
    if (b[i] >= b[i + 1] || i == n - 1) {
      m = i + 1;
      break;
    }
  }
  fill(b, b + n, 0);
  for (int i = 0; i < n; i++) {
    a[i].first = mp[a[i].first];
    b[a[i].first]++;
  }
  for (long long i = 0; i < m; i++) {
    y = 1;
    for (long long j = 1; j <= k; j++) {
      (((y += (dp[j - 1]) % mod) %= mod) *= b[i]) %= mod;
      v[i].push_back(y % mod);
    }
    for (long long j = 1; j <= k; j++) {
      (dp[j] += (v[i][j - 1] % mod)) %= mod;
    }
  }
  e = l / n;
  for (long long i = 1; i <= k; i++) {
    if (e < i) break;
    ans += ((dp[i] % mod) *
            (((e - i + 1) % mod) + ((e - i + 1) % mod < 0) * mod)) %
           mod;
    ans %= mod;
  }
  if (l % n != 0) {
    for (long long i = 0; i < n; i++) {
      if (a[i].second < l % n) {
        h[a[i].first]++;
      }
    }
    fill(dp, dp + k + 10, 0);
    for (long long i = 0; i < m; i++) {
      dp[0] = 1;
      for (long long j = 1; j <= k; j++) {
        (dp[j] += (v[i][j - 1] % mod)) %= mod;
      }
      for (long long j = 1; j <= k; j++) {
        if (e >= j - 1) (ans += dp[j - 1] * h[i]) %= mod;
      }
    }
  }
  cout << ans;
  return 0;
}

```