#include <bits/stdc++.h>
using namespace std;
int sum[1000000 + 5];
int A[1000000 + 5];
int dp[1000000 + 5][3];
int sdp[1000000 + 5][3];
int main() {
  int N, K;
  signed long long int L;
  map<signed long long int, int> mp;
  cin >> N >> L >> K;
  vector<int> v;
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
    v.push_back(A[i]);
  }
  sort(v.begin(), v.end());
  for (int i = 1; i <= N; ++i) {
    mp[v[i - 1]] = i;
  }
  for (int i = 1; i <= N; ++i) {
    A[i] = mp[A[i]];
    sum[i] = 1;
  }
  signed long long int ans = 0;
  for (int k = 1; k <= K; ++k) {
    for (int i = 1; i <= N; ++i) {
      dp[i][k & 1] = sum[A[i]];
      sdp[i][k & 1] = sdp[i][(k - 1) & 1] + dp[i][k & 1];
      if (sdp[i][k & 1] >= 1000000007) {
        sdp[i][k & 1] -= 1000000007;
      }
      if ((k - 1) * N + i <= L) {
        ans += sdp[i][k & 1];
        if (ans >= 1000000007) {
          ans -= 1000000007;
        }
      }
      signed long long int used = N * (K - 1) + i;
      signed long long int rem = (L - used) / N;
      if (rem > 0) {
        rem %= 1000000007;
        ans += (dp[i][k & 1] * rem) % 1000000007;
      }
      if (ans >= 1000000007) {
        ans -= 1000000007;
      }
    }
    for (int i = 1; i <= N; ++i) {
      sum[i] = 0;
    }
    for (int i = 1; i <= N; ++i) {
      sum[A[i]] += dp[i][k & 1];
      if (sum[A[i]] >= 1000000007) {
        sum[A[i]] -= 1000000007;
      }
    }
    for (int i = 1; i <= N; ++i) {
      sum[i] += sum[i - 1];
      if (sum[i] >= 1000000007) {
        sum[i] -= 1000000007;
      }
    }
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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;
int sum[1000000 + 5];
int A[1000000 + 5];
int dp[1000000 + 5][3];
int sdp[1000000 + 5][3];
int main() {
  int N, K;
  signed long long int L;
  map<signed long long int, int> mp;
  cin >> N >> L >> K;
  vector<int> v;
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
    v.push_back(A[i]);
  }
  sort(v.begin(), v.end());
  for (int i = 1; i <= N; ++i) {
    mp[v[i - 1]] = i;
  }
  for (int i = 1; i <= N; ++i) {
    A[i] = mp[A[i]];
    sum[i] = 1;
  }
  signed long long int ans = 0;
  for (int k = 1; k <= K; ++k) {
    for (int i = 1; i <= N; ++i) {
      dp[i][k & 1] = sum[A[i]];
      sdp[i][k & 1] = sdp[i][(k - 1) & 1] + dp[i][k & 1];
      if (sdp[i][k & 1] >= 1000000007) {
        sdp[i][k & 1] -= 1000000007;
      }
      if ((k - 1) * N + i <= L) {
        ans += sdp[i][k & 1];
        if (ans >= 1000000007) {
          ans -= 1000000007;
        }
      }
      signed long long int used = N * (K - 1) + i;
      signed long long int rem = (L - used) / N;
      if (rem > 0) {
        rem %= 1000000007;
        ans += (dp[i][k & 1] * rem) % 1000000007;
      }
      if (ans >= 1000000007) {
        ans -= 1000000007;
      }
    }
    for (int i = 1; i <= N; ++i) {
      sum[i] = 0;
    }
    for (int i = 1; i <= N; ++i) {
      sum[A[i]] += dp[i][k & 1];
      if (sum[A[i]] >= 1000000007) {
        sum[A[i]] -= 1000000007;
      }
    }
    for (int i = 1; i <= N; ++i) {
      sum[i] += sum[i - 1];
      if (sum[i] >= 1000000007) {
        sum[i] -= 1000000007;
      }
    }
  }
  cout << ans;
}

```