#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll const mod = 1000000000 + 7;
int a[1000000 + 5];
vector<pair<int, int> > vii;
vector<long long> dp[1000000 + 5];
int main() {
  int i, j, k;
  long long N, K;
  long long L, c, last;
  cin >> N >> L >> K;
  c = L / N;
  if (L % N) c++;
  last = (L - 1) % N;
  K = (int)min(c, (ll)K);
  for (i = 0; i < N; i++) {
    scanf("%d", &a[i]);
    vii.push_back(make_pair(a[i], i));
  }
  sort(vii.begin(), vii.end());
  for (i = 0; i <= N; i++) dp[i].resize(K + 1, 0);
  for (i = 0; i < N; i++) dp[i][1] = 1;
  for (j = 2; j <= K; j++) {
    ll sum = 0;
    int p = 0;
    for (i = 0; i < N; i++) {
      int id = vii[i].second;
      while (p < N && vii[p].first <= a[id])
        sum = (sum + dp[vii[p++].second][j - 1]) % mod;
      dp[id][j] = sum;
    }
  }
  ll res = 0;
  for (i = 0; i < N; i++)
    for (j = 1; j <= K; j++) {
      long long x = L / N;
      if (i <= last) x = c;
      x = x - j + 1;
      if (x > 0) res = (res + (x % mod) * dp[i][j]) % mod;
    }
  cout << res << endl;
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
ll const mod = 1000000000 + 7;
int a[1000000 + 5];
vector<pair<int, int> > vii;
vector<long long> dp[1000000 + 5];
int main() {
  int i, j, k;
  long long N, K;
  long long L, c, last;
  cin >> N >> L >> K;
  c = L / N;
  if (L % N) c++;
  last = (L - 1) % N;
  K = (int)min(c, (ll)K);
  for (i = 0; i < N; i++) {
    scanf("%d", &a[i]);
    vii.push_back(make_pair(a[i], i));
  }
  sort(vii.begin(), vii.end());
  for (i = 0; i <= N; i++) dp[i].resize(K + 1, 0);
  for (i = 0; i < N; i++) dp[i][1] = 1;
  for (j = 2; j <= K; j++) {
    ll sum = 0;
    int p = 0;
    for (i = 0; i < N; i++) {
      int id = vii[i].second;
      while (p < N && vii[p].first <= a[id])
        sum = (sum + dp[vii[p++].second][j - 1]) % mod;
      dp[id][j] = sum;
    }
  }
  ll res = 0;
  for (i = 0; i < N; i++)
    for (j = 1; j <= K; j++) {
      long long x = L / N;
      if (i <= last) x = c;
      x = x - j + 1;
      if (x > 0) res = (res + (x % mod) * dp[i][j]) % mod;
    }
  cout << res << endl;
  return 0;
}

```