#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
void solve() {
  long long int n, l, k, l1, left;
  cin >> n >> l >> k;
  l1 = l;
  l1 /= n;
  left = l % n;
  long long int A1[n];
  set<long long int> pop;
  set<long long int>::iterator it;
  map<long long int, long long int> lol;
  for (int i = 0; i < n; i++) {
    cin >> A1[i];
    pop.insert(A1[i]);
    lol[A1[i]]++;
  }
  vector<int> love, A;
  for (int i = 0; i < left; i++) {
    love.push_back(A1[i]);
  }
  sort(love.begin(), love.end());
  for (it = pop.begin(); it != pop.end(); it++) {
    A.push_back(*it);
  }
  long long int n1 = A.size();
  long long int dp[n1][k + 1];
  for (int i = 0; i < n1; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i][j] = 0;
    }
  }
  for (int i = 0; i < n1; i++) {
    dp[i][1] = lol[A[i]];
  }
  for (int i = 2; i <= k; i++) {
    long long int sum = 0;
    for (int j = 0; j < n1; j++) {
      sum += dp[j][i - 1];
      dp[j][i] += (lol[A[j]] * sum) % mod;
      sum %= mod;
      dp[j][i] %= mod;
    }
  }
  long long int ans = l;
  ans %= mod;
  for (int i = 0; i < n1; i++) {
    for (int j = 1; j <= k; j++) {
      if (j != 1) ans += (dp[i][j] * ((max(0LL, l1 - j + 1)) % mod)) % mod;
      ans %= mod;
      if (j != k && dp[i][j] != 0 && l1 - j + 1 > 0) {
        long long int kyu = love.size(), pp;
        pp = lower_bound(love.begin(), love.end(), A[i]) - love.begin();
        ans += (dp[i][j] * (kyu - pp)) % mod;
        ans %= mod;
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) solve();
}

### 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;
long long int mod = 1e9 + 7;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
void solve() {
  long long int n, l, k, l1, left;
  cin >> n >> l >> k;
  l1 = l;
  l1 /= n;
  left = l % n;
  long long int A1[n];
  set<long long int> pop;
  set<long long int>::iterator it;
  map<long long int, long long int> lol;
  for (int i = 0; i < n; i++) {
    cin >> A1[i];
    pop.insert(A1[i]);
    lol[A1[i]]++;
  }
  vector<int> love, A;
  for (int i = 0; i < left; i++) {
    love.push_back(A1[i]);
  }
  sort(love.begin(), love.end());
  for (it = pop.begin(); it != pop.end(); it++) {
    A.push_back(*it);
  }
  long long int n1 = A.size();
  long long int dp[n1][k + 1];
  for (int i = 0; i < n1; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i][j] = 0;
    }
  }
  for (int i = 0; i < n1; i++) {
    dp[i][1] = lol[A[i]];
  }
  for (int i = 2; i <= k; i++) {
    long long int sum = 0;
    for (int j = 0; j < n1; j++) {
      sum += dp[j][i - 1];
      dp[j][i] += (lol[A[j]] * sum) % mod;
      sum %= mod;
      dp[j][i] %= mod;
    }
  }
  long long int ans = l;
  ans %= mod;
  for (int i = 0; i < n1; i++) {
    for (int j = 1; j <= k; j++) {
      if (j != 1) ans += (dp[i][j] * ((max(0LL, l1 - j + 1)) % mod)) % mod;
      ans %= mod;
      if (j != k && dp[i][j] != 0 && l1 - j + 1 > 0) {
        long long int kyu = love.size(), pp;
        pp = lower_bound(love.begin(), love.end(), A[i]) - love.begin();
        ans += (dp[i][j] * (kyu - pp)) % mod;
        ans %= mod;
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) solve();
}

```