#include <bits/stdc++.h>
using namespace std;
pair<int, int> a[1100000];
int f[1100000];
vector<vector<int>> sum;
vector<vector<int>> dp;
const int mod = 1000000007;
const int m = mod;
int n, k;
long long l;
long long r[1100000];
int main() {
  scanf("%d %lld %d", &n, &l, &k);
  r[1] = 1;
  for (int i = 2; i < 1100000; ++i) r[i] = (m - (m / i) * r[m % i] % m) % m;
  dp = vector<vector<int>>(k + 2, vector<int>(n + 2, 0));
  sum = vector<vector<int>>(k + 2, vector<int>(n + 2, 0));
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = n - 1; i >= 0; i--) {
    f[i] = i;
    if (i != n - 1 && a[i].first == a[i + 1].first) f[i] = f[i + 1];
  }
  int ans = 0;
  long long full_runs = l / n;
  long long add_run = l % n;
  long long comb_full_i = 1;
  long long comb_full_i1 = 1;
  for (int(i) = 0; (i) < (int)(k + 1); (i)++) {
    for (int(j) = 0; (j) < (int)(n); (j)++) {
      if (i == 0) {
        if (j == 0)
          dp[i][j] = 1;
        else
          dp[i][j] = 0;
        continue;
      }
      dp[i][j] = sum[i - 1][f[j]];
      if (i <= full_runs) {
        long long choices = (full_runs - i + 1) % mod;
        ans = (ans + choices * dp[i][j]) % mod;
      }
      if (i <= full_runs + 1 && a[j].second < add_run) {
        long long choices = 1;
        ans = (ans + choices * dp[i][j]) % mod;
      }
    }
    for (int(j) = 0; (j) < (int)(n); (j)++) {
      if (j == 0)
        sum[i][j] = dp[i][j];
      else
        sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod;
    }
    comb_full_i1 = comb_full_i;
    comb_full_i = (comb_full_i * ((full_runs - i) % mod)) % mod;
    comb_full_i = (comb_full_i * r[i + 1]) % mod;
  }
  printf("%d\n", ans);
}

### Prompt

In cpp, your task is to solve 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;
pair<int, int> a[1100000];
int f[1100000];
vector<vector<int>> sum;
vector<vector<int>> dp;
const int mod = 1000000007;
const int m = mod;
int n, k;
long long l;
long long r[1100000];
int main() {
  scanf("%d %lld %d", &n, &l, &k);
  r[1] = 1;
  for (int i = 2; i < 1100000; ++i) r[i] = (m - (m / i) * r[m % i] % m) % m;
  dp = vector<vector<int>>(k + 2, vector<int>(n + 2, 0));
  sum = vector<vector<int>>(k + 2, vector<int>(n + 2, 0));
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = n - 1; i >= 0; i--) {
    f[i] = i;
    if (i != n - 1 && a[i].first == a[i + 1].first) f[i] = f[i + 1];
  }
  int ans = 0;
  long long full_runs = l / n;
  long long add_run = l % n;
  long long comb_full_i = 1;
  long long comb_full_i1 = 1;
  for (int(i) = 0; (i) < (int)(k + 1); (i)++) {
    for (int(j) = 0; (j) < (int)(n); (j)++) {
      if (i == 0) {
        if (j == 0)
          dp[i][j] = 1;
        else
          dp[i][j] = 0;
        continue;
      }
      dp[i][j] = sum[i - 1][f[j]];
      if (i <= full_runs) {
        long long choices = (full_runs - i + 1) % mod;
        ans = (ans + choices * dp[i][j]) % mod;
      }
      if (i <= full_runs + 1 && a[j].second < add_run) {
        long long choices = 1;
        ans = (ans + choices * dp[i][j]) % mod;
      }
    }
    for (int(j) = 0; (j) < (int)(n); (j)++) {
      if (j == 0)
        sum[i][j] = dp[i][j];
      else
        sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod;
    }
    comb_full_i1 = comb_full_i;
    comb_full_i = (comb_full_i * ((full_runs - i) % mod)) % mod;
    comb_full_i = (comb_full_i * r[i + 1]) % mod;
  }
  printf("%d\n", ans);
}

```