#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long n, l, k;
vector<long long> dp[1000100];
vector<long long> dp1[1000100];
int a[1000100];
int b[1000100];
int c[1000100];
map<int, int> m;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (int i = 1; i <= n; i++) {
    dp[i].resize(k + 1);
    dp1[i].resize(k + 1);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i - 1]);
    b[i - 1] = a[i - 1];
    dp[i][1] = 1;
  }
  sort(b, b + n);
  int top = 0;
  int temp1 = b[0];
  int top1 = 0;
  m[b[0]] = ++top;
  for (int i = 0; i < n; i++) {
    if (b[i] == temp1) {
      m[b[i]] = top;
      top1++;
    } else {
      c[top] = top1;
      m[b[i]] = ++top;
      top1 = 1;
      temp1 = b[i];
    }
  }
  c[top] = top1;
  for (int i = 1; i <= top; i++) dp[i][1] = 1;
  for (int i = 2; i <= k; i++) {
    long long temp = 0;
    for (int j = top; j >= 1; j--) {
      dp[j][i] = (temp + c[j] * dp[j][i - 1]) % mod;
      temp = (temp + c[j] * dp[j][i - 1]) % mod;
    }
  }
  for (int i = 1; i <= top; i++) dp1[i][1] = 1;
  for (int i = 2; i <= k; i++) {
    long long temp = 0;
    for (int j = 1; j <= top; j++) {
      dp1[j][i] = (temp + c[j] * dp1[j][i - 1]) % mod;
      temp = (temp + c[j] * dp1[j][i - 1]) % mod;
    }
  }
  long long ans = 0;
  long long now = (l / n);
  long long now1 = (l) % n;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      long long temp = m[a[i]];
      ans =
          (ans + dp[temp][j] * ((max(1ll * 0, now - j + 1)) % mod) % mod) % mod;
    }
  }
  for (int i = 0; i < now1; i++) {
    for (int j = 1; j <= k; j++) {
      if (j == 1) {
        ans = (ans + 1) % mod;
        continue;
      }
      int temp = m[a[i]];
      if (now >= j - 1) ans = (ans + dp1[temp][j]) % mod;
    }
  }
  printf("%lld\n", ans);
  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 long long mod = 1e9 + 7;
long long n, l, k;
vector<long long> dp[1000100];
vector<long long> dp1[1000100];
int a[1000100];
int b[1000100];
int c[1000100];
map<int, int> m;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  for (int i = 1; i <= n; i++) {
    dp[i].resize(k + 1);
    dp1[i].resize(k + 1);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i - 1]);
    b[i - 1] = a[i - 1];
    dp[i][1] = 1;
  }
  sort(b, b + n);
  int top = 0;
  int temp1 = b[0];
  int top1 = 0;
  m[b[0]] = ++top;
  for (int i = 0; i < n; i++) {
    if (b[i] == temp1) {
      m[b[i]] = top;
      top1++;
    } else {
      c[top] = top1;
      m[b[i]] = ++top;
      top1 = 1;
      temp1 = b[i];
    }
  }
  c[top] = top1;
  for (int i = 1; i <= top; i++) dp[i][1] = 1;
  for (int i = 2; i <= k; i++) {
    long long temp = 0;
    for (int j = top; j >= 1; j--) {
      dp[j][i] = (temp + c[j] * dp[j][i - 1]) % mod;
      temp = (temp + c[j] * dp[j][i - 1]) % mod;
    }
  }
  for (int i = 1; i <= top; i++) dp1[i][1] = 1;
  for (int i = 2; i <= k; i++) {
    long long temp = 0;
    for (int j = 1; j <= top; j++) {
      dp1[j][i] = (temp + c[j] * dp1[j][i - 1]) % mod;
      temp = (temp + c[j] * dp1[j][i - 1]) % mod;
    }
  }
  long long ans = 0;
  long long now = (l / n);
  long long now1 = (l) % n;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= k; j++) {
      long long temp = m[a[i]];
      ans =
          (ans + dp[temp][j] * ((max(1ll * 0, now - j + 1)) % mod) % mod) % mod;
    }
  }
  for (int i = 0; i < now1; i++) {
    for (int j = 1; j <= k; j++) {
      if (j == 1) {
        ans = (ans + 1) % mod;
        continue;
      }
      int temp = m[a[i]];
      if (now >= j - 1) ans = (ans + dp1[temp][j]) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```