#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int NUM = 1e6 + 9;
long long n, l, k;
int arr[NUM];
long long dp[5 * NUM];
struct node {
  long long val, pos;
} number[NUM];
bool cmp(node a, node b) {
  if (a.val != b.val) {
    return a.val < b.val;
  } else {
    return a.pos < b.pos;
  }
}
int main() {
  while (scanf("%I64d%I64d%I64d", &n, &l, &k) != EOF) {
    for (long long i = 0; i < n; i++) {
      scanf("%I64d", &arr[i]);
    }
    long long s = n * k;
    for (long long i = 0; i < s; i++) {
      number[i].val = arr[i % n];
      number[i].pos = i;
    }
    sort(number, number + s, cmp);
    long long ans = l;
    for (long long i = 0; i < s; i++) {
      long long pos = number[i].pos;
      if (pos < n) {
        dp[0] = (dp[0] + 1) % MOD;
      } else if (pos < l) {
        dp[pos / n] = (dp[pos / n - 1] + dp[pos / n]) % MOD;
      }
      if (pos < l) {
        if (pos >= n) {
          long long m = (l - pos - 1 + n) / n;
          ans = (ans + m % MOD * dp[pos / n - 1]) % MOD;
        }
      }
    }
    cout << ans % MOD << endl;
  }
  return 0;
}

### 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;
const int MOD = 1e9 + 7;
const int NUM = 1e6 + 9;
long long n, l, k;
int arr[NUM];
long long dp[5 * NUM];
struct node {
  long long val, pos;
} number[NUM];
bool cmp(node a, node b) {
  if (a.val != b.val) {
    return a.val < b.val;
  } else {
    return a.pos < b.pos;
  }
}
int main() {
  while (scanf("%I64d%I64d%I64d", &n, &l, &k) != EOF) {
    for (long long i = 0; i < n; i++) {
      scanf("%I64d", &arr[i]);
    }
    long long s = n * k;
    for (long long i = 0; i < s; i++) {
      number[i].val = arr[i % n];
      number[i].pos = i;
    }
    sort(number, number + s, cmp);
    long long ans = l;
    for (long long i = 0; i < s; i++) {
      long long pos = number[i].pos;
      if (pos < n) {
        dp[0] = (dp[0] + 1) % MOD;
      } else if (pos < l) {
        dp[pos / n] = (dp[pos / n - 1] + dp[pos / n]) % MOD;
      }
      if (pos < l) {
        if (pos >= n) {
          long long m = (l - pos - 1 + n) / n;
          ans = (ans + m % MOD * dp[pos / n - 1]) % MOD;
        }
      }
    }
    cout << ans % MOD << endl;
  }
  return 0;
}

```