#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const long long LINF = 1LL << 50;
const int M = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
int seq[M], seq2[M];
long long dp[2][M];
const int MOD = 1e9 + 7;
int main() {
  long long n, l, k;
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &seq[i]);
    seq2[i] = seq[i];
  }
  sort(seq2, seq2 + n);
  for (int i = 0; i < n; i++)
    seq[i] = lower_bound(seq2, seq2 + n, seq[i]) - seq2;
  int maxn = 0;
  for (int i = 0; i < n; i++) maxn = max(maxn, seq[i]);
  maxn++;
  long long c = l / n, d = l % n;
  for (int i = 0; i < d; i++) seq2[i] = seq[i];
  long long ans = 0;
  int id = 0;
  for (int i = 0; i < min(c, k); i++) {
    id ^= 1;
    for (int j = 0; j <= maxn; j++) dp[id][j] = 0;
    for (int j = 0; j < n; j++)
      dp[id][seq[j]] = (dp[id][seq[j]] + dp[id ^ 1][seq[j]] + 1) % MOD;
    for (int j = 1; j < maxn; j++)
      dp[id][j] = (dp[id][j] + dp[id][j - 1]) % MOD;
    ans = (ans + dp[id][maxn - 1]) % MOD;
  }
  if (c > k) ans = (ans + dp[id][maxn - 1] * ((c - k) % MOD)) % MOD;
  int t = id;
  if (c >= k) t ^= 1;
  for (int i = 0; i < d; i++) {
    ans += dp[t][seq2[i]] + 1;
    ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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 int INF = 1 << 30;
const long long LINF = 1LL << 50;
const int M = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-6;
int seq[M], seq2[M];
long long dp[2][M];
const int MOD = 1e9 + 7;
int main() {
  long long n, l, k;
  scanf("%I64d %I64d %I64d", &n, &l, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &seq[i]);
    seq2[i] = seq[i];
  }
  sort(seq2, seq2 + n);
  for (int i = 0; i < n; i++)
    seq[i] = lower_bound(seq2, seq2 + n, seq[i]) - seq2;
  int maxn = 0;
  for (int i = 0; i < n; i++) maxn = max(maxn, seq[i]);
  maxn++;
  long long c = l / n, d = l % n;
  for (int i = 0; i < d; i++) seq2[i] = seq[i];
  long long ans = 0;
  int id = 0;
  for (int i = 0; i < min(c, k); i++) {
    id ^= 1;
    for (int j = 0; j <= maxn; j++) dp[id][j] = 0;
    for (int j = 0; j < n; j++)
      dp[id][seq[j]] = (dp[id][seq[j]] + dp[id ^ 1][seq[j]] + 1) % MOD;
    for (int j = 1; j < maxn; j++)
      dp[id][j] = (dp[id][j] + dp[id][j - 1]) % MOD;
    ans = (ans + dp[id][maxn - 1]) % MOD;
  }
  if (c > k) ans = (ans + dp[id][maxn - 1] * ((c - k) % MOD)) % MOD;
  int t = id;
  if (c >= k) t ^= 1;
  for (int i = 0; i < d; i++) {
    ans += dp[t][seq2[i]] + 1;
    ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}

```