#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
struct mint {
  int value;
  mint() : value(0) {}
  template <class T>
  mint(T x) {
    value = x % M;
    if (value < 0) {
      value += M;
    }
  }
  mint &operator++() {
    if (++value == M) {
      value = 0;
    }
    return *this;
  }
  mint &operator--() {
    if (--value == -1) {
      value = M - 1;
    }
    return *this;
  }
  mint operator++(int) {
    mint res = *this;
    ++*this;
    return res;
  }
  mint operator--(int) {
    mint res = *this;
    --*this;
    return res;
  }
  mint &operator+=(mint oth) {
    value += oth.value;
    if (value >= M) {
      value -= M;
    }
    return *this;
  }
  mint &operator-=(mint oth) {
    value -= oth.value;
    if (value < 0) {
      value += M;
    }
    return *this;
  }
  mint &operator*=(mint oth) {
    value = (long long)value * oth.value % M;
    return *this;
  }
  mint operator+() const { return *this; }
  mint operator-() const { return mint() - *this; }
  mint pow(long long e = M - 2) const {
    mint res = 1, b = *this;
    while (e > 0) {
      if (e % 2 == 1) {
        res *= b;
      }
      b *= b;
      e /= 2;
    }
    return res;
  }
  friend mint operator+(mint a, mint b) { return a += b; }
  friend mint operator-(mint a, mint b) { return a -= b; }
  friend mint operator*(mint a, mint b) { return mint(a) *= b; }
  friend bool operator==(mint a, mint b) { return a.value == b.value; }
  friend bool operator!=(mint a, mint b) { return a.value != b.value; }
  friend ostream &operator<<(ostream &out, mint a) {
    out << a.value;
    return out;
  }
};
const int N = 1000000 + 1;
map<int, mint> ways[N];
int a[N], v[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    v[i] = a[i];
    ways[0][v[i]] = 1;
  }
  sort(v, v + n);
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      if (ways[i].count(v[j]) == 0) {
        ways[i][v[j]] = (j == 0) ? 0 : ways[i][v[j - 1]];
      }
      ways[i][v[j]] += ways[i - 1][v[j]];
    }
  }
  mint ans = 0;
  for (int i = 1; i <= k; ++i) {
    ans += max(0LL, l / n + 1 - i) * ways[i][v[n - 1]];
    if (i - 1 <= l / n) {
      for (int j = 0; j < l % n; ++j) {
        ans += ways[i - 1][a[j]];
      }
    }
  }
  cout << ans << "\n";
}

### 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 M = 1000000007;
struct mint {
  int value;
  mint() : value(0) {}
  template <class T>
  mint(T x) {
    value = x % M;
    if (value < 0) {
      value += M;
    }
  }
  mint &operator++() {
    if (++value == M) {
      value = 0;
    }
    return *this;
  }
  mint &operator--() {
    if (--value == -1) {
      value = M - 1;
    }
    return *this;
  }
  mint operator++(int) {
    mint res = *this;
    ++*this;
    return res;
  }
  mint operator--(int) {
    mint res = *this;
    --*this;
    return res;
  }
  mint &operator+=(mint oth) {
    value += oth.value;
    if (value >= M) {
      value -= M;
    }
    return *this;
  }
  mint &operator-=(mint oth) {
    value -= oth.value;
    if (value < 0) {
      value += M;
    }
    return *this;
  }
  mint &operator*=(mint oth) {
    value = (long long)value * oth.value % M;
    return *this;
  }
  mint operator+() const { return *this; }
  mint operator-() const { return mint() - *this; }
  mint pow(long long e = M - 2) const {
    mint res = 1, b = *this;
    while (e > 0) {
      if (e % 2 == 1) {
        res *= b;
      }
      b *= b;
      e /= 2;
    }
    return res;
  }
  friend mint operator+(mint a, mint b) { return a += b; }
  friend mint operator-(mint a, mint b) { return a -= b; }
  friend mint operator*(mint a, mint b) { return mint(a) *= b; }
  friend bool operator==(mint a, mint b) { return a.value == b.value; }
  friend bool operator!=(mint a, mint b) { return a.value != b.value; }
  friend ostream &operator<<(ostream &out, mint a) {
    out << a.value;
    return out;
  }
};
const int N = 1000000 + 1;
map<int, mint> ways[N];
int a[N], v[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  long long l;
  cin >> n >> l >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    v[i] = a[i];
    ways[0][v[i]] = 1;
  }
  sort(v, v + n);
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      if (ways[i].count(v[j]) == 0) {
        ways[i][v[j]] = (j == 0) ? 0 : ways[i][v[j - 1]];
      }
      ways[i][v[j]] += ways[i - 1][v[j]];
    }
  }
  mint ans = 0;
  for (int i = 1; i <= k; ++i) {
    ans += max(0LL, l / n + 1 - i) * ways[i][v[n - 1]];
    if (i - 1 <= l / n) {
      for (int j = 0; j < l % n; ++j) {
        ans += ways[i - 1][a[j]];
      }
    }
  }
  cout << ans << "\n";
}

```