#include <bits/stdc++.h>
int N, K, dp[1000001], a[1000001], b[1000001], cnt[1000001], rank[1000001], ans;
long long L;
int binSearch(int k) {
  int l = 0, r = N - 1, mid;
  while (l < r) {
    mid = l + r >> 1;
    if (b[mid] < k)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
int main() {
  int i, j, r;
  long long q;
  scanf("%d%lld%d", &N, &L, &K);
  for (i = 0; i < N; ++i) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  std::sort(b, b + N);
  for (i = 1; i < N; ++i) rank[i] = rank[i - 1] + (b[i] != b[i - 1]);
  for (i = 0; i < N; ++i) {
    a[i] = rank[binSearch(a[i])];
    ++cnt[a[i]];
  }
  q = L / N;
  r = L % N;
  K = std::min((long long)K, q + (r > 0));
  for (i = 0; i <= rank[N - 1]; ++i) dp[i] = 1;
  for (i = 1; i <= K; ++i) {
    for (j = 0; j < r; ++j) {
      ans += dp[a[j]];
      if (ans >= 1000000007) ans -= 1000000007;
    }
    for (j = 0; j <= rank[N - 1]; ++j)
      dp[j] = (long long)dp[j] * cnt[j] % 1000000007;
    for (j = 1; j <= rank[N - 1]; ++j) {
      dp[j] += dp[j - 1];
      if (dp[j] >= 1000000007) dp[j] -= 1000000007;
    }
    ans = (ans + (q - i + 1) % 1000000007 * dp[rank[N - 1]]) % 1000000007;
  }
  printf("%d", 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>
int N, K, dp[1000001], a[1000001], b[1000001], cnt[1000001], rank[1000001], ans;
long long L;
int binSearch(int k) {
  int l = 0, r = N - 1, mid;
  while (l < r) {
    mid = l + r >> 1;
    if (b[mid] < k)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
int main() {
  int i, j, r;
  long long q;
  scanf("%d%lld%d", &N, &L, &K);
  for (i = 0; i < N; ++i) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  std::sort(b, b + N);
  for (i = 1; i < N; ++i) rank[i] = rank[i - 1] + (b[i] != b[i - 1]);
  for (i = 0; i < N; ++i) {
    a[i] = rank[binSearch(a[i])];
    ++cnt[a[i]];
  }
  q = L / N;
  r = L % N;
  K = std::min((long long)K, q + (r > 0));
  for (i = 0; i <= rank[N - 1]; ++i) dp[i] = 1;
  for (i = 1; i <= K; ++i) {
    for (j = 0; j < r; ++j) {
      ans += dp[a[j]];
      if (ans >= 1000000007) ans -= 1000000007;
    }
    for (j = 0; j <= rank[N - 1]; ++j)
      dp[j] = (long long)dp[j] * cnt[j] % 1000000007;
    for (j = 1; j <= rank[N - 1]; ++j) {
      dp[j] += dp[j - 1];
      if (dp[j] >= 1000000007) dp[j] -= 1000000007;
    }
    ans = (ans + (q - i + 1) % 1000000007 * dp[rank[N - 1]]) % 1000000007;
  }
  printf("%d", ans);
  return 0;
}

```