#include <bits/stdc++.h>
using namespace std;
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return (a % b == 0 ? b : GCD(b, a % b));
}
long long POW(long long base, long long exp) {
  long long val;
  val = 1;
  while (exp > 0) {
    if (exp % 2 == 1) {
      val = (val * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    exp = exp / 2;
  }
  return val;
}
vector<pair<long long, long long> > a;
vector<vector<long long> > dp;
int main() {
  long long n, k, l, r, s, x, i, j, it, sum, ans, num;
  scanf("%lld%lld%lld", &n, &l, &k);
  r = l / n + 1;
  r = r % 1000000007;
  s = (r - 1 + 1000000007) % 1000000007;
  x = l % n;
  for (i = 0; i < n; i++) {
    scanf("%lld", &num);
    a.push_back(make_pair(num, i));
  }
  sort(a.begin(), a.end());
  dp.resize(k + 1);
  for (i = 0; i <= k; i++) {
    dp[i].resize(n);
  }
  ans = 0;
  for (i = 0; i < n; i++) {
    dp[1][i] = 1;
  }
  for (i = 0; i < n; i++) {
    if (a[i].second < x) {
      ans = (ans + r * dp[1][i]) % 1000000007;
    } else {
      ans = (ans + s * dp[1][i]) % 1000000007;
    }
  }
  for (i = 2; i <= min(k, l / n + 1); i++) {
    sum = 0;
    for (j = 0; j < n;) {
      it = j;
      while (it < n && a[it].first == a[j].first) {
        sum = (sum + dp[i - 1][it]) % 1000000007;
        it++;
      }
      it = j;
      while (it < n && a[it].first == a[j].first) {
        dp[i][it] = sum;
        it++;
      }
      j = it;
    }
    for (j = 0; j < n; j++) {
      if (a[j].second < x) {
        ans = (ans + (r - i + 1 + 1000000007) * dp[i][j]) % 1000000007;
      } else {
        ans = (ans + (s - i + 1 + 1000000007) * dp[i][j]) % 1000000007;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 GCD(long long a, long long b) {
  if (b == 0) return a;
  return (a % b == 0 ? b : GCD(b, a % b));
}
long long POW(long long base, long long exp) {
  long long val;
  val = 1;
  while (exp > 0) {
    if (exp % 2 == 1) {
      val = (val * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    exp = exp / 2;
  }
  return val;
}
vector<pair<long long, long long> > a;
vector<vector<long long> > dp;
int main() {
  long long n, k, l, r, s, x, i, j, it, sum, ans, num;
  scanf("%lld%lld%lld", &n, &l, &k);
  r = l / n + 1;
  r = r % 1000000007;
  s = (r - 1 + 1000000007) % 1000000007;
  x = l % n;
  for (i = 0; i < n; i++) {
    scanf("%lld", &num);
    a.push_back(make_pair(num, i));
  }
  sort(a.begin(), a.end());
  dp.resize(k + 1);
  for (i = 0; i <= k; i++) {
    dp[i].resize(n);
  }
  ans = 0;
  for (i = 0; i < n; i++) {
    dp[1][i] = 1;
  }
  for (i = 0; i < n; i++) {
    if (a[i].second < x) {
      ans = (ans + r * dp[1][i]) % 1000000007;
    } else {
      ans = (ans + s * dp[1][i]) % 1000000007;
    }
  }
  for (i = 2; i <= min(k, l / n + 1); i++) {
    sum = 0;
    for (j = 0; j < n;) {
      it = j;
      while (it < n && a[it].first == a[j].first) {
        sum = (sum + dp[i - 1][it]) % 1000000007;
        it++;
      }
      it = j;
      while (it < n && a[it].first == a[j].first) {
        dp[i][it] = sum;
        it++;
      }
      j = it;
    }
    for (j = 0; j < n; j++) {
      if (a[j].second < x) {
        ans = (ans + (r - i + 1 + 1000000007) * dp[i][j]) % 1000000007;
      } else {
        ans = (ans + (s - i + 1 + 1000000007) * dp[i][j]) % 1000000007;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```