#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
long long n, m, k, l;
long long DP[(1000000) + 2];
long long cnt[(1000000) + 2], cnt1[(1000000) + 2];
long long A[(1000000) + 2];
pair<long long, long long> C[(1000000) + 2];
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  long long l1 = l / n, l2 = l % n;
  for (int i = 1; i <= n; ++i) scanf("%I64d", A + i);
  m = n * k;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= k; ++j)
      C[i + (j - 1) * n] = pair<long long, long long>(A[i], i + (j - 1) * n);
  sort(C + 1, C + m + 1);
  for (int i = 1; i <= m; ++i) {
    long long id = C[i].second;
    if (id <= n)
      DP[i] = 1;
    else
      DP[i] = (cnt[(id - 1) / n] + 1) % (1000000007);
    cnt[(id - 1) / n + 1] = (cnt[(id - 1) / n + 1] + DP[i]) % (1000000007);
    if ((id - 1) % n + 1 <= l2)
      cnt1[(id - 1) / n + 1] = (cnt1[(id - 1) / n + 1] + DP[i]) % (1000000007);
  }
  long long ans = 0;
  for (int i = 1; i <= k; ++i) {
    if (l1 - i < 0) break;
    ans = (ans + (cnt[i] - cnt[i - 1]) * ((l1 - i + 1) % (1000000007))) %
          (1000000007);
  }
  if (l2) ans += cnt1[min(l1 + 1, k)];
  cout << (ans % (1000000007) + (1000000007)) % (1000000007);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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>
#pragma GCC optimize("O2")
using namespace std;
long long n, m, k, l;
long long DP[(1000000) + 2];
long long cnt[(1000000) + 2], cnt1[(1000000) + 2];
long long A[(1000000) + 2];
pair<long long, long long> C[(1000000) + 2];
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  long long l1 = l / n, l2 = l % n;
  for (int i = 1; i <= n; ++i) scanf("%I64d", A + i);
  m = n * k;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= k; ++j)
      C[i + (j - 1) * n] = pair<long long, long long>(A[i], i + (j - 1) * n);
  sort(C + 1, C + m + 1);
  for (int i = 1; i <= m; ++i) {
    long long id = C[i].second;
    if (id <= n)
      DP[i] = 1;
    else
      DP[i] = (cnt[(id - 1) / n] + 1) % (1000000007);
    cnt[(id - 1) / n + 1] = (cnt[(id - 1) / n + 1] + DP[i]) % (1000000007);
    if ((id - 1) % n + 1 <= l2)
      cnt1[(id - 1) / n + 1] = (cnt1[(id - 1) / n + 1] + DP[i]) % (1000000007);
  }
  long long ans = 0;
  for (int i = 1; i <= k; ++i) {
    if (l1 - i < 0) break;
    ans = (ans + (cnt[i] - cnt[i - 1]) * ((l1 - i + 1) % (1000000007))) %
          (1000000007);
  }
  if (l2) ans += cnt1[min(l1 + 1, k)];
  cout << (ans % (1000000007) + (1000000007)) % (1000000007);
  return 0;
}

```