#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
const long long int md = 1e9 + 7;
long long int n, k, l, dp[2][MAXN], prs[2][MAXN], a[MAXN], ans = 0, m;
map<int, int> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  m = l / n;
  set<int> st;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    st.insert(a[i]);
  }
  int pt = 0;
  while (!st.empty()) mp[*st.begin()] = pt, pt++, st.erase(*st.begin());
  ans = l % md;
  for (int i = 0; i < n; i++) dp[0][i] = 1ll, prs[0][mp[a[i]]]++;
  for (int i = 1; i < pt; i++) prs[0][i] += prs[0][i - 1], prs[0][i] %= md;
  for (int i = 1; i < k; i++) {
    int ii = i % 2, i2 = (i + 1) % 2;
    for (int j = 0; j < pt; j++) prs[ii][j] = 0;
    for (int j = 0; j < n; j++) {
      dp[ii][j] = prs[i2][mp[a[j]]],
      prs[ii][mp[a[j]]] = (prs[ii][mp[a[j]]] + dp[ii][j]) % md;
      long long int tmp = (max(m - i, 0ll) + (l % n > j && m >= i)) % md;
      ans += ((tmp * dp[ii][j]) % md);
      ans %= md;
    }
    for (int j = 0; j < pt; j++)
      prs[ii][j] = (prs[ii][j] + (j > 0 ? prs[ii][j - 1] : 0ll)) % md;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 MAXN = 1e6 + 10;
const long long int md = 1e9 + 7;
long long int n, k, l, dp[2][MAXN], prs[2][MAXN], a[MAXN], ans = 0, m;
map<int, int> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l >> k;
  m = l / n;
  set<int> st;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    st.insert(a[i]);
  }
  int pt = 0;
  while (!st.empty()) mp[*st.begin()] = pt, pt++, st.erase(*st.begin());
  ans = l % md;
  for (int i = 0; i < n; i++) dp[0][i] = 1ll, prs[0][mp[a[i]]]++;
  for (int i = 1; i < pt; i++) prs[0][i] += prs[0][i - 1], prs[0][i] %= md;
  for (int i = 1; i < k; i++) {
    int ii = i % 2, i2 = (i + 1) % 2;
    for (int j = 0; j < pt; j++) prs[ii][j] = 0;
    for (int j = 0; j < n; j++) {
      dp[ii][j] = prs[i2][mp[a[j]]],
      prs[ii][mp[a[j]]] = (prs[ii][mp[a[j]]] + dp[ii][j]) % md;
      long long int tmp = (max(m - i, 0ll) + (l % n > j && m >= i)) % md;
      ans += ((tmp * dp[ii][j]) % md);
      ans %= md;
    }
    for (int j = 0; j < pt; j++)
      prs[ii][j] = (prs[ii][j] + (j > 0 ? prs[ii][j - 1] : 0ll)) % md;
  }
  cout << ans << endl;
  return 0;
}

```