#include <bits/stdc++.h>
using namespace std;
const long long inf = 1000000000000000000LL;
const long long mod = 1e9 + 7;
long long add(long long a, long long b) { return (a + b) % mod; }
long long mul(long long a, long long b) { return (long long)a * b % mod; }
const int MAXN = 1000005;
int n, k;
long long l;
int a[MAXN];
void pref_sum(map<long long, long long> &m) {
  long long sum = 0;
  for (auto &it : m) {
    sum = add(sum, it.second);
    it.second = sum;
  }
}
int main(void) {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = (0); i < (n); ++i) scanf("%d", a + i);
  long long B = l / n;
  map<long long, long long> cnt;
  for (int i = (0); i < (n); ++i) ++cnt[a[i]];
  map<long long, long long> seq_cnt = cnt;
  long long sol = l % mod;
  for (int i = (0); i < (k - 1); ++i) {
    pref_sum(seq_cnt);
    long long count;
    if (l % n) {
      count = 0;
      for (int j = (0); j < (l % n); ++j) {
        count = add(count, seq_cnt[a[j]]);
      }
      if (B >= i + 1) sol = add(sol, count);
    }
    count = 0;
    for (auto it : cnt) {
      seq_cnt[it.first] = mul(seq_cnt[it.first], it.second);
      count = add(count, seq_cnt[it.first]);
    }
    if (B >= i + 2) sol = add(sol, mul(count, (B - i - 1) % mod));
  }
  printf("%lld\n", sol);
  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 long long inf = 1000000000000000000LL;
const long long mod = 1e9 + 7;
long long add(long long a, long long b) { return (a + b) % mod; }
long long mul(long long a, long long b) { return (long long)a * b % mod; }
const int MAXN = 1000005;
int n, k;
long long l;
int a[MAXN];
void pref_sum(map<long long, long long> &m) {
  long long sum = 0;
  for (auto &it : m) {
    sum = add(sum, it.second);
    it.second = sum;
  }
}
int main(void) {
  scanf("%d%lld%d", &n, &l, &k);
  for (int i = (0); i < (n); ++i) scanf("%d", a + i);
  long long B = l / n;
  map<long long, long long> cnt;
  for (int i = (0); i < (n); ++i) ++cnt[a[i]];
  map<long long, long long> seq_cnt = cnt;
  long long sol = l % mod;
  for (int i = (0); i < (k - 1); ++i) {
    pref_sum(seq_cnt);
    long long count;
    if (l % n) {
      count = 0;
      for (int j = (0); j < (l % n); ++j) {
        count = add(count, seq_cnt[a[j]]);
      }
      if (B >= i + 1) sol = add(sol, count);
    }
    count = 0;
    for (auto it : cnt) {
      seq_cnt[it.first] = mul(seq_cnt[it.first], it.second);
      count = add(count, seq_cnt[it.first]);
    }
    if (B >= i + 2) sol = add(sol, mul(count, (B - i - 1) % mod));
  }
  printf("%lld\n", sol);
  return 0;
}

```