#include <bits/stdc++.h>
using namespace std;
int M = 1000000007;
int a[1000005];
int rnk[1000005];
bool comp(int i, int j) { return a[i] < a[j]; }
int main() {
  long long n, k;
  long long l;
  cin >> n >> l >> k;
  long long r = l % n;
  l = l / n;
  vector<pair<int, int> > aa;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    aa.push_back(make_pair(a[i], i));
  }
  sort(aa.begin(), aa.end());
  for (int i = n - 1; i >= 0; i--) {
    if (i < n - 1 && aa[i].first == aa[i + 1].first)
      rnk[aa[i].second] = rnk[aa[i + 1].second];
    else
      rnk[aa[i].second] = i;
  }
  vector<long long> dp(n, 1);
  long long ans = (l * n + r) % M;
  k = min(k, l + 1);
  for (int x = 2; x <= k; x++) {
    vector<long long> dp2(n, 0);
    vector<long long> sdp(n, 0);
    for (int i = 0; i < n; i++) {
      sdp[rnk[i]] += dp[i];
      sdp[rnk[i]] %= M;
    }
    for (int i = 1; i < n; i++) {
      sdp[i] += sdp[i - 1];
      sdp[i] %= M;
    }
    for (int i = 0; i < n; i++) {
      dp2[i] = sdp[rnk[i]];
      if (i < r)
        ans += dp2[i] * ((l - x + 2) % M);
      else
        ans += dp2[i] * ((l - x + 1) % M);
      ans %= M;
    }
    dp = dp2;
  }
  cout << ans;
  return 0;
}

### 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;
int M = 1000000007;
int a[1000005];
int rnk[1000005];
bool comp(int i, int j) { return a[i] < a[j]; }
int main() {
  long long n, k;
  long long l;
  cin >> n >> l >> k;
  long long r = l % n;
  l = l / n;
  vector<pair<int, int> > aa;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    aa.push_back(make_pair(a[i], i));
  }
  sort(aa.begin(), aa.end());
  for (int i = n - 1; i >= 0; i--) {
    if (i < n - 1 && aa[i].first == aa[i + 1].first)
      rnk[aa[i].second] = rnk[aa[i + 1].second];
    else
      rnk[aa[i].second] = i;
  }
  vector<long long> dp(n, 1);
  long long ans = (l * n + r) % M;
  k = min(k, l + 1);
  for (int x = 2; x <= k; x++) {
    vector<long long> dp2(n, 0);
    vector<long long> sdp(n, 0);
    for (int i = 0; i < n; i++) {
      sdp[rnk[i]] += dp[i];
      sdp[rnk[i]] %= M;
    }
    for (int i = 1; i < n; i++) {
      sdp[i] += sdp[i - 1];
      sdp[i] %= M;
    }
    for (int i = 0; i < n; i++) {
      dp2[i] = sdp[rnk[i]];
      if (i < r)
        ans += dp2[i] * ((l - x + 2) % M);
      else
        ans += dp2[i] * ((l - x + 1) % M);
      ans %= M;
    }
    dp = dp2;
  }
  cout << ans;
  return 0;
}

```