#include <bits/stdc++.h>
using ll = long long;
const ll mod = 1e9 + 7;
const ll N = 1e6 + 5;
const int base = 400;
using ull = unsigned long long;
using namespace std;
ll n, m, t, k, T, a[N], fe[N], ans, st[N], u, v, b[N];
vector<ll> adj[65], res, kq;
vector<vector<ll>> dp;
pair<ll, ll> p[N];
void sol() {
  cin >> n >> m >> k;
  dp.resize(n + 2);
  for (int i = 0; i < n; i++) {
    cin >> p[i].first;
    p[i].second = i;
    dp[i].resize(k + 2);
    dp[i][0] = 1;
    t = (m + n - 1) / n;
    u = m % n;
    if (u == 0) u = n;
    if (u <= i) --t;
    ans = (ans + t) % mod;
  }
  sort(p, p + n);
  k = min(k, (m + n - 1) / n);
  for (int i = 1; i < k; i++) {
    int id = 0;
    ll total = 0;
    for (int j = 0; j < n; j++) {
      while (id < n && p[id].first <= p[j].first) {
        total = (total + dp[id][i - 1]) % mod;
        ++id;
      }
      dp[j][i] = total;
      t = (m + n - 1) / n;
      u = m % n;
      if (u == 0) u = n;
      if (u <= p[j].second) --t;
      t = (t - i) % mod;
      ans = (ans + t * dp[j][i] % mod) % mod;
    }
  }
  cout << ans;
}
int main() {
  if (fopen("pieaters"
            ".inp",
            "r")) {
    freopen(
        "pieaters"
        ".inp",
        "r", stdin);
    freopen(
        "pieaters"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  ntest = 1;
  while (ntest-- > 0) sol();
}

### 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 ll = long long;
const ll mod = 1e9 + 7;
const ll N = 1e6 + 5;
const int base = 400;
using ull = unsigned long long;
using namespace std;
ll n, m, t, k, T, a[N], fe[N], ans, st[N], u, v, b[N];
vector<ll> adj[65], res, kq;
vector<vector<ll>> dp;
pair<ll, ll> p[N];
void sol() {
  cin >> n >> m >> k;
  dp.resize(n + 2);
  for (int i = 0; i < n; i++) {
    cin >> p[i].first;
    p[i].second = i;
    dp[i].resize(k + 2);
    dp[i][0] = 1;
    t = (m + n - 1) / n;
    u = m % n;
    if (u == 0) u = n;
    if (u <= i) --t;
    ans = (ans + t) % mod;
  }
  sort(p, p + n);
  k = min(k, (m + n - 1) / n);
  for (int i = 1; i < k; i++) {
    int id = 0;
    ll total = 0;
    for (int j = 0; j < n; j++) {
      while (id < n && p[id].first <= p[j].first) {
        total = (total + dp[id][i - 1]) % mod;
        ++id;
      }
      dp[j][i] = total;
      t = (m + n - 1) / n;
      u = m % n;
      if (u == 0) u = n;
      if (u <= p[j].second) --t;
      t = (t - i) % mod;
      ans = (ans + t * dp[j][i] % mod) % mod;
    }
  }
  cout << ans;
}
int main() {
  if (fopen("pieaters"
            ".inp",
            "r")) {
    freopen(
        "pieaters"
        ".inp",
        "r", stdin);
    freopen(
        "pieaters"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  ntest = 1;
  while (ntest-- > 0) sol();
}

```