#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int maxn = 1e6 + 6;
long long n, l, k;
int value[maxn], dp[maxn];
pair<int, int> pii[maxn];
int main() {
  while (~scanf("%I64d%I64d%I64d", &n, &l, &k)) {
    memset(dp, 0, sizeof dp);
    for (int i = 0; i < n; i++) {
      scanf("%d", value + i);
    }
    for (int i = 0; i < n * k; i++) {
      pii[i] = make_pair(value[i % n], i);
    }
    sort(pii, pii + n * k);
    long long pre = 1, ans = 0;
    for (int i = 0; i < n * k; i++) {
      int idx = pii[i].second;
      if (idx < n) {
        pre = 1;
      } else {
        pre = dp[idx / n - 1];
      }
      dp[idx / n] = (dp[idx / n] + pre) % MOD;
      if (idx < l) {
        long long cnt = (l - idx - 1 + n) / n % MOD;
        ans = (ans + (long long)pre * cnt % MOD) % MOD;
      }
    }
    printf("%I64d\n", (ans % MOD + MOD) % MOD);
  }
  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 MOD = 1e9 + 7;
const int maxn = 1e6 + 6;
long long n, l, k;
int value[maxn], dp[maxn];
pair<int, int> pii[maxn];
int main() {
  while (~scanf("%I64d%I64d%I64d", &n, &l, &k)) {
    memset(dp, 0, sizeof dp);
    for (int i = 0; i < n; i++) {
      scanf("%d", value + i);
    }
    for (int i = 0; i < n * k; i++) {
      pii[i] = make_pair(value[i % n], i);
    }
    sort(pii, pii + n * k);
    long long pre = 1, ans = 0;
    for (int i = 0; i < n * k; i++) {
      int idx = pii[i].second;
      if (idx < n) {
        pre = 1;
      } else {
        pre = dp[idx / n - 1];
      }
      dp[idx / n] = (dp[idx / n] + pre) % MOD;
      if (idx < l) {
        long long cnt = (l - idx - 1 + n) / n % MOD;
        ans = (ans + (long long)pre * cnt % MOD) % MOD;
      }
    }
    printf("%I64d\n", (ans % MOD + MOD) % MOD);
  }
  return 0;
}

```