#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long max_n = 1e6 + 20;
long long n, l, k, maxi;
map<long long, long long> mp, cnt;
map<pair<long long, long long>, long long> dp;
vector<long long> vec, sec;
long long save[max_n];
long long pw(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    b <<= 1;
    a = (a * a) % MOD;
  }
  return ret;
}
void init() {
  save[1] = 1;
  for (long long i = 2; i < max_n; i++) save[i] = (save[i - 1] * i) % MOD;
  for (long long i = 1; i <= k; i++)
    for (long long j = 0; j < sec.size(); j++) {
      if (i == 1) {
        dp[{i, sec[j]}] = cnt[sec[j]];
        if (j != 0)
          dp[{i, sec[j]}] = (dp[{i, sec[j]}] + dp[{i, sec[j - 1]}]) % MOD;
        continue;
      } else {
        if (j == 0)
          dp[{i, sec[j]}] = (cnt[sec[0]] * dp[{i - 1, sec[j]}]) % MOD;
        else
          dp[{i, sec[j]}] =
              (cnt[sec[j]] * dp[{i - 1, sec[j]}] + dp[{i, sec[j - 1]}]) % MOD;
      }
    }
}
long long cal(long long v) {
  if (l < (v - 1) * n) return 0;
  l -= (v - 1) * n;
  long long z = l / n;
  z %= MOD;
  long long ret = (z * dp[{v, maxi}]) % MOD;
  for (long long i = 0; i < l % n; i++)
    if (v != 1)
      ret = (ret + dp[{v - 1, vec[i]}]) % MOD;
    else
      ret = (ret + 1) % MOD;
  return ret;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  vec.resize(n);
  for (long long i = 0; i < n; i++) {
    scanf("%I64d", &vec[i]);
    cnt[vec[i]]++;
    maxi = max(maxi, vec[i]);
  }
  sec.assign(vec.begin(), vec.end());
  sort(sec.begin(), sec.end());
  sec.resize(unique(sec.begin(), sec.end()) - sec.begin());
  init();
  long long ans = 0;
  long long secl = l;
  for (long long i = 1; i <= k; i++) {
    l = secl;
    ans = (ans + cal(i)) % MOD;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate 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 MOD = 1e9 + 7;
const long long max_n = 1e6 + 20;
long long n, l, k, maxi;
map<long long, long long> mp, cnt;
map<pair<long long, long long>, long long> dp;
vector<long long> vec, sec;
long long save[max_n];
long long pw(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    b <<= 1;
    a = (a * a) % MOD;
  }
  return ret;
}
void init() {
  save[1] = 1;
  for (long long i = 2; i < max_n; i++) save[i] = (save[i - 1] * i) % MOD;
  for (long long i = 1; i <= k; i++)
    for (long long j = 0; j < sec.size(); j++) {
      if (i == 1) {
        dp[{i, sec[j]}] = cnt[sec[j]];
        if (j != 0)
          dp[{i, sec[j]}] = (dp[{i, sec[j]}] + dp[{i, sec[j - 1]}]) % MOD;
        continue;
      } else {
        if (j == 0)
          dp[{i, sec[j]}] = (cnt[sec[0]] * dp[{i - 1, sec[j]}]) % MOD;
        else
          dp[{i, sec[j]}] =
              (cnt[sec[j]] * dp[{i - 1, sec[j]}] + dp[{i, sec[j - 1]}]) % MOD;
      }
    }
}
long long cal(long long v) {
  if (l < (v - 1) * n) return 0;
  l -= (v - 1) * n;
  long long z = l / n;
  z %= MOD;
  long long ret = (z * dp[{v, maxi}]) % MOD;
  for (long long i = 0; i < l % n; i++)
    if (v != 1)
      ret = (ret + dp[{v - 1, vec[i]}]) % MOD;
    else
      ret = (ret + 1) % MOD;
  return ret;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  vec.resize(n);
  for (long long i = 0; i < n; i++) {
    scanf("%I64d", &vec[i]);
    cnt[vec[i]]++;
    maxi = max(maxi, vec[i]);
  }
  sec.assign(vec.begin(), vec.end());
  sort(sec.begin(), sec.end());
  sec.resize(unique(sec.begin(), sec.end()) - sec.begin());
  init();
  long long ans = 0;
  long long secl = l;
  for (long long i = 1; i <= k; i++) {
    l = secl;
    ans = (ans + cal(i)) % MOD;
  }
  printf("%I64d\n", ans);
  return 0;
}

```