#include <bits/stdc++.h>
using namespace std;
const int N = 1100000;
const int mod = 1e9 + 7;
int a[N], b[N], cnt[N], sum[N];
vector<vector<int>> dp;
void add(int &u, int v) {
  u += v;
  if (u >= mod) u -= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) b[i - 1] = a[i];
  sort(b, b + n);
  int m = unique(b, b + n) - b;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b, b + m, a[i]) - b + 1;
  dp.resize(k + 1);
  for (int i = 0; i <= k; i++) dp[i].resize(m + 1);
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= m; j++) dp[i][j] = 0;
  for (int i = 1; i <= n; i++) cnt[a[i]]++;
  for (int i = 1; i <= n; i++) dp[1][a[i]]++;
  for (int i = 1; i <= m; i++) add(dp[1][i], dp[1][i - 1]);
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= m; j++)
      dp[i][j] = (long long)dp[i - 1][j] * cnt[j] % mod;
    for (int j = 1; j <= m; j++) add(dp[i][j], dp[i][j - 1]);
  }
  long long how = l / n;
  int ans = 0;
  for (int i = 1; i <= k; i++) {
    if (how - i + 1 >= 0) {
      add(ans, (long long)dp[i][m] * ((how - i + 1) % mod) % mod);
    }
  }
  for (int i = 1; i <= l % n; i++) {
    for (int j = 1; j <= min<long long>(k - 1, how); j++) {
      add(ans, dp[j][a[i]]);
    }
    add(ans, 1);
  }
  cout << ans << endl;
  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>
using namespace std;
const int N = 1100000;
const int mod = 1e9 + 7;
int a[N], b[N], cnt[N], sum[N];
vector<vector<int>> dp;
void add(int &u, int v) {
  u += v;
  if (u >= mod) u -= mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) b[i - 1] = a[i];
  sort(b, b + n);
  int m = unique(b, b + n) - b;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b, b + m, a[i]) - b + 1;
  dp.resize(k + 1);
  for (int i = 0; i <= k; i++) dp[i].resize(m + 1);
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= m; j++) dp[i][j] = 0;
  for (int i = 1; i <= n; i++) cnt[a[i]]++;
  for (int i = 1; i <= n; i++) dp[1][a[i]]++;
  for (int i = 1; i <= m; i++) add(dp[1][i], dp[1][i - 1]);
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= m; j++)
      dp[i][j] = (long long)dp[i - 1][j] * cnt[j] % mod;
    for (int j = 1; j <= m; j++) add(dp[i][j], dp[i][j - 1]);
  }
  long long how = l / n;
  int ans = 0;
  for (int i = 1; i <= k; i++) {
    if (how - i + 1 >= 0) {
      add(ans, (long long)dp[i][m] * ((how - i + 1) % mod) % mod);
    }
  }
  for (int i = 1; i <= l % n; i++) {
    for (int j = 1; j <= min<long long>(k - 1, how); j++) {
      add(ans, dp[j][a[i]]);
    }
    add(ans, 1);
  }
  cout << ans << endl;
  return 0;
}

```