#include <bits/stdc++.h>
using namespace std;
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
long long dp[1000005], a[1000005], sum[1000005], ans;
vector<int> v;
int main() {
  int n, k, N;
  long long l;
  scanf("%d %lld %d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  N = v.size();
  sum[0] = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 1; j <= N; j++) sum[j] = 0LL;
    for (int j = 0; j < n; j++) sum[a[j]] = (sum[a[j]] + dp[j]) % 1000000007;
    for (int j = 1; j <= N; j++) sum[j] = (sum[j - 1] + sum[j]) % 1000000007;
    sum[0] = 0;
    for (int j = 0; j < n; j++) {
      dp[j] = sum[a[j]];
      if (l / n + (l % n > j ? 1 : 0) - i <= 0) break;
      ans = (ans + dp[j] * ((l / n + (l % n > j ? 1 : 0) - i) % 1000000007)) %
            1000000007;
    }
  }
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in CPP 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;
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
long long dp[1000005], a[1000005], sum[1000005], ans;
vector<int> v;
int main() {
  int n, k, N;
  long long l;
  scanf("%d %lld %d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  N = v.size();
  sum[0] = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 1; j <= N; j++) sum[j] = 0LL;
    for (int j = 0; j < n; j++) sum[a[j]] = (sum[a[j]] + dp[j]) % 1000000007;
    for (int j = 1; j <= N; j++) sum[j] = (sum[j - 1] + sum[j]) % 1000000007;
    sum[0] = 0;
    for (int j = 0; j < n; j++) {
      dp[j] = sum[a[j]];
      if (l / n + (l % n > j ? 1 : 0) - i <= 0) break;
      ans = (ans + dp[j] * ((l / n + (l % n > j ? 1 : 0) - i) % 1000000007)) %
            1000000007;
    }
  }
  printf("%lld\n", ans);
}

```