#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
const int nn = (N - 10) / 2;
const int M = 1000000007;
int n, k, a[N];
long long l;
vector<int> v;
long long times(int p, int k) {
  int pl = l % n;
  if (pl == 0) pl = n;
  int t = int(p <= pl);
  long long ts = (l - 1) / n + t;
  return max(0LL, ts - k + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> l >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; ++i)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  vector<vector<int> > dp(k + 5);
  vector<vector<int> > sm(k + 5);
  for (int i = 1; i <= k; ++i) dp[i].resize(n + 5);
  for (int i = 1; i <= k; ++i) sm[i].resize(n + 5);
  for (int j = 1; j <= n; ++j) {
    dp[1][j] = 1;
    sm[1][a[j]]++;
  }
  for (int j = 1; j <= n; ++j) sm[1][j] += sm[1][j - 1];
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      dp[i][j] = sm[i - 1][a[j]] % M;
    }
    for (int j = 1; j <= n; ++j) {
      sm[i][a[j]] += dp[i][j];
      sm[i][a[j]] %= M;
    }
    for (int j = 1; j <= n; ++j) {
      sm[i][j] += sm[i][j - 1];
      sm[i][j] %= M;
    }
  }
  int ans = 0;
  for (int kk = 1; kk <= k; ++kk) {
    for (int i = 1; i <= n; ++i)
      ans = (ans + (1LL * dp[kk][i] * (times(i, kk) % M)) % M) % M;
  }
  cout << ans << '\n';
}

### 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;
const int N = 1000010;
const int nn = (N - 10) / 2;
const int M = 1000000007;
int n, k, a[N];
long long l;
vector<int> v;
long long times(int p, int k) {
  int pl = l % n;
  if (pl == 0) pl = n;
  int t = int(p <= pl);
  long long ts = (l - 1) / n + t;
  return max(0LL, ts - k + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> l >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; ++i)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  vector<vector<int> > dp(k + 5);
  vector<vector<int> > sm(k + 5);
  for (int i = 1; i <= k; ++i) dp[i].resize(n + 5);
  for (int i = 1; i <= k; ++i) sm[i].resize(n + 5);
  for (int j = 1; j <= n; ++j) {
    dp[1][j] = 1;
    sm[1][a[j]]++;
  }
  for (int j = 1; j <= n; ++j) sm[1][j] += sm[1][j - 1];
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      dp[i][j] = sm[i - 1][a[j]] % M;
    }
    for (int j = 1; j <= n; ++j) {
      sm[i][a[j]] += dp[i][j];
      sm[i][a[j]] %= M;
    }
    for (int j = 1; j <= n; ++j) {
      sm[i][j] += sm[i][j - 1];
      sm[i][j] %= M;
    }
  }
  int ans = 0;
  for (int kk = 1; kk <= k; ++kk) {
    for (int i = 1; i <= n; ++i)
      ans = (ans + (1LL * dp[kk][i] * (times(i, kk) % M)) % M) % M;
  }
  cout << ans << '\n';
}

```