#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e6 + 10, MOD = 1000 * 1000 * 1000 + 7;
long long dp[MAXN], a[MAXN], b[MAXN], cnt[MAXN], ind[MAXN];
map<long long, long long> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k, l, m = 0;
  cin >> n >> l >> k;
  for (long long i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++) ind[i] = upper_bound(b, b + n, a[i]) - b - 1;
  for (long long i = 0; i < k; i++) {
    for (long long j = 0; j < n; j++) {
      long long index = i * n + j;
      dp[index] = i == 0 ? 1 : cnt[ind[j]];
      dp[index] %= MOD;
    }
    for (long long j = 0; j < n; j++) cnt[j] = 0;
    for (long long j = 0; j < n; j++) cnt[ind[j]] += (dp[i * n + j]);
    long long las = 0;
    for (long long j = 0; j < n; j++) {
      cnt[j] += las;
      cnt[j] %= MOD;
      las = cnt[j];
    }
  }
  long long sum = 0;
  long long q = l % n;
  long long mul = l / n;
  for (long long i = 0; i < k; i++) {
    if (mul < 0) break;
    for (long long j = 0; j < n; j++) {
      long long MUL = mul % MOD;
      sum += (dp[i * n + j] * MUL) % MOD;
      sum %= MOD;
      if (j < q) {
        sum += (dp[i * n + j]) % MOD;
        sum %= MOD;
      }
    }
    --mul;
  }
  cout << sum << endl;
}

### 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 long long MAXN = 1e6 + 10, MOD = 1000 * 1000 * 1000 + 7;
long long dp[MAXN], a[MAXN], b[MAXN], cnt[MAXN], ind[MAXN];
map<long long, long long> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k, l, m = 0;
  cin >> n >> l >> k;
  for (long long i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++) ind[i] = upper_bound(b, b + n, a[i]) - b - 1;
  for (long long i = 0; i < k; i++) {
    for (long long j = 0; j < n; j++) {
      long long index = i * n + j;
      dp[index] = i == 0 ? 1 : cnt[ind[j]];
      dp[index] %= MOD;
    }
    for (long long j = 0; j < n; j++) cnt[j] = 0;
    for (long long j = 0; j < n; j++) cnt[ind[j]] += (dp[i * n + j]);
    long long las = 0;
    for (long long j = 0; j < n; j++) {
      cnt[j] += las;
      cnt[j] %= MOD;
      las = cnt[j];
    }
  }
  long long sum = 0;
  long long q = l % n;
  long long mul = l / n;
  for (long long i = 0; i < k; i++) {
    if (mul < 0) break;
    for (long long j = 0; j < n; j++) {
      long long MUL = mul % MOD;
      sum += (dp[i * n + j] * MUL) % MOD;
      sum %= MOD;
      if (j < q) {
        sum += (dp[i * n + j]) % MOD;
        sum %= MOD;
      }
    }
    --mul;
  }
  cout << sum << endl;
}

```