#include <bits/stdc++.h>
using namespace std;
const long long N = 1001000, INF = 1e18, MOD = 1000000007;
long long n, l, k, m, q = 1, a[N], b[N], f[N], prefsum[N], z, u, mx, mn, ans;
pair<long long, long long> p[N];
string s;
vector<int> v[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n >> l >> k;
  long long h = l % n, t = (((l + n - 1) / n) % MOD);
  k = min(k, (l + n - 1) / n);
  long long dp[k + 5][n + 5];
  if (!h) h = n;
  for (int i = 1; i <= n; i++)
    cin >> p[i].first, prefsum[i] = p[i].second = b[i] = dp[2][i] = i;
  sort(p + 1, p + n + 1);
  for (int i = n; i >= 1; i--)
    if (p[i + 1].first == p[i].first) dp[2][i] = b[i] = b[i + 1];
  for (int i = 2; i <= k; i++) {
    t--;
    if (!t) t += MOD;
    for (int j = 1; j <= n; j++)
      (ans += dp[i][j] * (t - (p[j].second > h))) %= MOD;
    for (int j = 1; j <= n; j++) prefsum[j] = (prefsum[j - 1] + dp[i][j]) % MOD;
    for (int j = 1; j <= n; j++) dp[i + 1][j] = prefsum[b[j]];
  }
  cout << (ans + l) % MOD;
}

### Prompt

Generate 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>
using namespace std;
const long long N = 1001000, INF = 1e18, MOD = 1000000007;
long long n, l, k, m, q = 1, a[N], b[N], f[N], prefsum[N], z, u, mx, mn, ans;
pair<long long, long long> p[N];
string s;
vector<int> v[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n >> l >> k;
  long long h = l % n, t = (((l + n - 1) / n) % MOD);
  k = min(k, (l + n - 1) / n);
  long long dp[k + 5][n + 5];
  if (!h) h = n;
  for (int i = 1; i <= n; i++)
    cin >> p[i].first, prefsum[i] = p[i].second = b[i] = dp[2][i] = i;
  sort(p + 1, p + n + 1);
  for (int i = n; i >= 1; i--)
    if (p[i + 1].first == p[i].first) dp[2][i] = b[i] = b[i + 1];
  for (int i = 2; i <= k; i++) {
    t--;
    if (!t) t += MOD;
    for (int j = 1; j <= n; j++)
      (ans += dp[i][j] * (t - (p[j].second > h))) %= MOD;
    for (int j = 1; j <= n; j++) prefsum[j] = (prefsum[j - 1] + dp[i][j]) % MOD;
    for (int j = 1; j <= n; j++) dp[i + 1][j] = prefsum[b[j]];
  }
  cout << (ans + l) % MOD;
}

```