#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
const int M = 100;
const int inf = (int)1e9 + 7;
const long long base = (long long)1e18;
const double pi = acos(-1);
const double ep = 1e-18;
int a[N];
int n, k;
long long l;
long long lt[N], invlt[N];
int sl[N];
vector<int> b;
vector<int> dp[N];
vector<int> sum[N];
long long cnt[N];
long long preT[N];
long long pw(int x, int k) {
  if (k == 0) return 1;
  long long tmp = pw(x, k / 2);
  tmp = tmp * tmp % inf;
  if (k & 1) tmp = tmp * x % inf;
  return tmp;
}
long long C(int v) {
  if (v == 0) return 1;
  return (long long)preT[v - 1] * invlt[v] % inf;
}
map<int, int> st;
map<int, int> rs;
int L[N];
void solve() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    st[a[i]]++;
  }
  int x = 0;
  for (map<int, int>::iterator it = st.begin(); it != st.end(); it++) {
    rs[it->first] = ++x;
    L[x] = it->second;
  }
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < k + 1; j++) {
      dp[i].push_back(0);
      sum[i].push_back(0);
    }
  }
  long long t = l / n;
  int r = l % n;
  for (int i = 1; i < n + 1; i++) {
    dp[i][0] = 1;
    sum[i][0] = 1;
    for (int j = 1; j < k + 1; j++) {
      dp[i][j] = (long long)L[i] * sum[i][j - 1] % inf;
      sum[i][j] = (sum[i - 1][j] + dp[i][j]) % inf;
      cnt[j] = (cnt[j] + dp[i][j]) % inf;
    }
  }
  long long ans = 0;
  for (int i = 1; i < k + 1; i++) {
    if (i > t) break;
    long long tmp = (long long)(t - i + 1) % inf * cnt[i] % inf;
    ans = (ans + tmp) % inf;
  }
  for (int i = 1; i < k + 1; i++) {
    if (i - 1 > t) break;
    for (int j = 0; j < r; j++) {
      long long tmp = sum[rs[a[j]]][i - 1] % inf;
      ans = (ans + tmp) % inf;
    }
  }
  cout << ans;
}
int main() {
  solve();
  return 0;
}

### Prompt

Your task is to create 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 int N = 1e6 + 7;
const int M = 100;
const int inf = (int)1e9 + 7;
const long long base = (long long)1e18;
const double pi = acos(-1);
const double ep = 1e-18;
int a[N];
int n, k;
long long l;
long long lt[N], invlt[N];
int sl[N];
vector<int> b;
vector<int> dp[N];
vector<int> sum[N];
long long cnt[N];
long long preT[N];
long long pw(int x, int k) {
  if (k == 0) return 1;
  long long tmp = pw(x, k / 2);
  tmp = tmp * tmp % inf;
  if (k & 1) tmp = tmp * x % inf;
  return tmp;
}
long long C(int v) {
  if (v == 0) return 1;
  return (long long)preT[v - 1] * invlt[v] % inf;
}
map<int, int> st;
map<int, int> rs;
int L[N];
void solve() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    st[a[i]]++;
  }
  int x = 0;
  for (map<int, int>::iterator it = st.begin(); it != st.end(); it++) {
    rs[it->first] = ++x;
    L[x] = it->second;
  }
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < k + 1; j++) {
      dp[i].push_back(0);
      sum[i].push_back(0);
    }
  }
  long long t = l / n;
  int r = l % n;
  for (int i = 1; i < n + 1; i++) {
    dp[i][0] = 1;
    sum[i][0] = 1;
    for (int j = 1; j < k + 1; j++) {
      dp[i][j] = (long long)L[i] * sum[i][j - 1] % inf;
      sum[i][j] = (sum[i - 1][j] + dp[i][j]) % inf;
      cnt[j] = (cnt[j] + dp[i][j]) % inf;
    }
  }
  long long ans = 0;
  for (int i = 1; i < k + 1; i++) {
    if (i > t) break;
    long long tmp = (long long)(t - i + 1) % inf * cnt[i] % inf;
    ans = (ans + tmp) % inf;
  }
  for (int i = 1; i < k + 1; i++) {
    if (i - 1 > t) break;
    for (int j = 0; j < r; j++) {
      long long tmp = sum[rs[a[j]]][i - 1] % inf;
      ans = (ans + tmp) % inf;
    }
  }
  cout << ans;
}
int main() {
  solve();
  return 0;
}

```