#include <bits/stdc++.h>
using namespace std;
const int maxn = static_cast<int>(1e6 + 10);
const int mod = static_cast<int>(1e9 + 7);
long long a[maxn], cnt[maxn], cnt2[maxn];
int main() {
  long long n, l, k;
  scanf("%lld%lld%lld", &n, &l, &k);
  vector<pair<long long, long long> > asort;
  vector<int> v;
  long long diff = l % n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    v.push_back(a[i]);
    asort.push_back(make_pair(a[i], i));
  }
  sort(v.begin(), v.end());
  sort(asort.begin(), asort.end());
  for (int i = 0; i < static_cast<int>((asort).size()); i++) {
    asort[i].first = i;
    swap(asort[i].first, asort[i].second);
  }
  sort(asort.begin(), asort.end());
  long long D = min(l / n + 1, k);
  vector<vector<long long> > dp(n + 2, vector<long long>(D + 2, 0));
  for (int i = 0; i < n; i++) {
    dp[i][1] = 1;
  }
  for (int j = 2; j <= D; j++) {
    long long pref = 0;
    int pos = 0;
    for (int i = 0; i < n; i++) {
      while (pos < n && v[pos] <= v[i]) {
        pref = (pref + dp[pos][j - 1]) % mod;
        pos++;
      }
      dp[i][j] = pref;
    }
  }
  long long answer = 0;
  long long e = l / n;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= D; j++) {
      answer += (dp[i][j] * ((e - j + 1) % mod)) % mod;
      while (answer >= mod) answer -= mod;
    }
  }
  long long T = min(k, e + 1);
  for (int i = 0; i < diff; i++) {
    for (int j = 1; j <= T; j++) {
      answer += dp[asort[i].second][j];
      while (answer >= mod) answer -= mod;
    }
  }
  printf("%lld\n", answer);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = static_cast<int>(1e6 + 10);
const int mod = static_cast<int>(1e9 + 7);
long long a[maxn], cnt[maxn], cnt2[maxn];
int main() {
  long long n, l, k;
  scanf("%lld%lld%lld", &n, &l, &k);
  vector<pair<long long, long long> > asort;
  vector<int> v;
  long long diff = l % n;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    v.push_back(a[i]);
    asort.push_back(make_pair(a[i], i));
  }
  sort(v.begin(), v.end());
  sort(asort.begin(), asort.end());
  for (int i = 0; i < static_cast<int>((asort).size()); i++) {
    asort[i].first = i;
    swap(asort[i].first, asort[i].second);
  }
  sort(asort.begin(), asort.end());
  long long D = min(l / n + 1, k);
  vector<vector<long long> > dp(n + 2, vector<long long>(D + 2, 0));
  for (int i = 0; i < n; i++) {
    dp[i][1] = 1;
  }
  for (int j = 2; j <= D; j++) {
    long long pref = 0;
    int pos = 0;
    for (int i = 0; i < n; i++) {
      while (pos < n && v[pos] <= v[i]) {
        pref = (pref + dp[pos][j - 1]) % mod;
        pos++;
      }
      dp[i][j] = pref;
    }
  }
  long long answer = 0;
  long long e = l / n;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= D; j++) {
      answer += (dp[i][j] * ((e - j + 1) % mod)) % mod;
      while (answer >= mod) answer -= mod;
    }
  }
  long long T = min(k, e + 1);
  for (int i = 0; i < diff; i++) {
    for (int j = 1; j <= T; j++) {
      answer += dp[asort[i].second][j];
      while (answer >= mod) answer -= mod;
    }
  }
  printf("%lld\n", answer);
  return 0;
}

```