#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, k;
long long l;
vector<int> dp[1000100];
int a[1000100];
int p[1000100];
int A[1000100];
bool cmp(int i, int j) { return a[i] < a[j]; }
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int sum[1000100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    dp[i].resize(k + 1, 0);
    dp[i][1] = 1;
    p[i] = i;
    A[i] = a[i];
  }
  sort(A, A + n);
  int nn = unique(A, A + n) - A;
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(A, A + nn, a[i]) - A;
  }
  for (int i = 1; i < k; ++i) {
    for (int j = 0; j < nn; ++j) {
      sum[j] = 0;
    }
    for (int j = 0; j < n; ++j) {
      add(sum[a[j]], dp[j][i]);
    }
    for (int j = 1; j < nn; ++j) {
      add(sum[j], sum[j - 1]);
    }
    for (int j = 0; j < n; ++j) {
      dp[j][i + 1] = sum[a[j]];
    }
  }
  int ans = l % mod;
  for (int i = 2; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      long long t = l / n;
      if (i > t)
        t = 0;
      else
        t = t - i + 1;
      add(ans, 1ll * t % mod * 1ll * dp[j][i] % mod);
    }
    if (i <= l / n + 1)
      for (int j = 0; j < l % n; ++j) {
        add(ans, dp[j][i]);
      }
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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 mod = 1e9 + 7;
int n, k;
long long l;
vector<int> dp[1000100];
int a[1000100];
int p[1000100];
int A[1000100];
bool cmp(int i, int j) { return a[i] < a[j]; }
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int sum[1000100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    dp[i].resize(k + 1, 0);
    dp[i][1] = 1;
    p[i] = i;
    A[i] = a[i];
  }
  sort(A, A + n);
  int nn = unique(A, A + n) - A;
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(A, A + nn, a[i]) - A;
  }
  for (int i = 1; i < k; ++i) {
    for (int j = 0; j < nn; ++j) {
      sum[j] = 0;
    }
    for (int j = 0; j < n; ++j) {
      add(sum[a[j]], dp[j][i]);
    }
    for (int j = 1; j < nn; ++j) {
      add(sum[j], sum[j - 1]);
    }
    for (int j = 0; j < n; ++j) {
      dp[j][i + 1] = sum[a[j]];
    }
  }
  int ans = l % mod;
  for (int i = 2; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      long long t = l / n;
      if (i > t)
        t = 0;
      else
        t = t - i + 1;
      add(ans, 1ll * t % mod * 1ll * dp[j][i] % mod);
    }
    if (i <= l / n + 1)
      for (int j = 0; j < l % n; ++j) {
        add(ans, dp[j][i]);
      }
  }
  cout << ans;
  return 0;
}

```