#include <bits/stdc++.h>
using namespace std;
const int N = 200;
typedef long long matrix[N][N];
const long long MOD = 1000000007;
const double EPS = 1e-8;
const double PI = 2 * acos(0);
const int maxn = 1e6 + 6;
long long n, l, k;
vector<vector<long long> > dp, g;
vector<int> v;
int a[1001000];
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  long long t = l / n;
  long long p = l % n;
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
  dp.resize(k + 3);
  g.resize(k + 3);
  for (int i = 0; i < k + 3; i++) {
    dp[i].resize(n + 3);
    g[i].resize(n + 3);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) g[0][i] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < n; j++) {
      dp[i][j] = g[i - 1][a[j]];
      long long d = t - i + 1;
      if (j < p) d++;
      if (d < 0LL) d = 0LL;
      ans = (ans + (d % MOD) * dp[i][j]) % MOD;
    }
    for (int j = 0; j < n; j++) g[i][a[j]] = (g[i][a[j]] + dp[i][j]) % MOD;
    for (int j = 1; j < n; j++) g[i][j] = (g[i][j] + g[i][j - 1]) % MOD;
  }
  cout << ans;
}

### 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 N = 200;
typedef long long matrix[N][N];
const long long MOD = 1000000007;
const double EPS = 1e-8;
const double PI = 2 * acos(0);
const int maxn = 1e6 + 6;
long long n, l, k;
vector<vector<long long> > dp, g;
vector<int> v;
int a[1001000];
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  long long t = l / n;
  long long p = l % n;
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
  dp.resize(k + 3);
  g.resize(k + 3);
  for (int i = 0; i < k + 3; i++) {
    dp[i].resize(n + 3);
    g[i].resize(n + 3);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) g[0][i] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < n; j++) {
      dp[i][j] = g[i - 1][a[j]];
      long long d = t - i + 1;
      if (j < p) d++;
      if (d < 0LL) d = 0LL;
      ans = (ans + (d % MOD) * dp[i][j]) % MOD;
    }
    for (int j = 0; j < n; j++) g[i][a[j]] = (g[i][a[j]] + dp[i][j]) % MOD;
    for (int j = 1; j < n; j++) g[i][j] = (g[i][j] + g[i][j - 1]) % MOD;
  }
  cout << ans;
}

```