#include <bits/stdc++.h>
#pragma comment(linker, "/STATK:134217728")
using namespace std;
const int MOD = (int)1e9 + 7;
int n, k;
long long L;
int a[1 << 20];
int t[1 << 20];
int cnt[1 << 20];
map<int, int> M;
long long f[1 << 20];
long long inv[1 << 20];
long long powmod(long long a, int x) {
  if (x == 0) return 1;
  if (x & 1) return a * powmod(a * a % MOD, x / 2) % MOD;
  return powmod(a * a % MOD, x / 2);
}
vector<vector<long long> > dp;
long long go(int pos, int k) {
  if (pos < 0) return k == 0;
  long long& res = dp[pos][k];
  if (res != -1) return res;
  res = go(pos - 1, k);
  if (k > 0) res = (res + cnt[pos] * go(pos, k - 1)) % MOD;
  return res;
}
int main() {
  f[0] = 1;
  for (int i = 1; i < 1 << 20; ++i) f[i] = f[i - 1] * i % MOD;
  for (int i = 0; i < 1 << 20; ++i) inv[i] = powmod(f[i], MOD - 2);
  scanf("%d%I64d%d", &n, &L, &k);
  long long have = L / n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    t[i] = a[i];
  }
  sort(t, t + n);
  for (int i = 0; i < n; ++i) M[t[i]] = i;
  for (int i = 0; i < n; ++i) a[i] = M[a[i]];
  for (int i = 0; i < n; ++i) cnt[a[i]]++;
  long long res = 0;
  dp = vector<vector<long long> >(n + 1, vector<long long>(k + 1, -1));
  for (int i = 1; i <= k; ++i) {
    if (i > have) break;
    long long mul = (have - i + 1) % MOD;
    res = (res + mul * go(n, i)) % MOD;
  }
  int m = L % n;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j <= k - 1; ++j) {
      if (j > have) break;
      res = (res + go(a[i], j)) % MOD;
    }
  }
  printf("%I64d\n", res);
  return 0;
}

### Prompt

Create a solution in CPP for 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>
#pragma comment(linker, "/STATK:134217728")
using namespace std;
const int MOD = (int)1e9 + 7;
int n, k;
long long L;
int a[1 << 20];
int t[1 << 20];
int cnt[1 << 20];
map<int, int> M;
long long f[1 << 20];
long long inv[1 << 20];
long long powmod(long long a, int x) {
  if (x == 0) return 1;
  if (x & 1) return a * powmod(a * a % MOD, x / 2) % MOD;
  return powmod(a * a % MOD, x / 2);
}
vector<vector<long long> > dp;
long long go(int pos, int k) {
  if (pos < 0) return k == 0;
  long long& res = dp[pos][k];
  if (res != -1) return res;
  res = go(pos - 1, k);
  if (k > 0) res = (res + cnt[pos] * go(pos, k - 1)) % MOD;
  return res;
}
int main() {
  f[0] = 1;
  for (int i = 1; i < 1 << 20; ++i) f[i] = f[i - 1] * i % MOD;
  for (int i = 0; i < 1 << 20; ++i) inv[i] = powmod(f[i], MOD - 2);
  scanf("%d%I64d%d", &n, &L, &k);
  long long have = L / n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    t[i] = a[i];
  }
  sort(t, t + n);
  for (int i = 0; i < n; ++i) M[t[i]] = i;
  for (int i = 0; i < n; ++i) a[i] = M[a[i]];
  for (int i = 0; i < n; ++i) cnt[a[i]]++;
  long long res = 0;
  dp = vector<vector<long long> >(n + 1, vector<long long>(k + 1, -1));
  for (int i = 1; i <= k; ++i) {
    if (i > have) break;
    long long mul = (have - i + 1) % MOD;
    res = (res + mul * go(n, i)) % MOD;
  }
  int m = L % n;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j <= k - 1; ++j) {
      if (j > have) break;
      res = (res + go(a[i], j)) % MOD;
    }
  }
  printf("%I64d\n", res);
  return 0;
}

```