#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int RAND() { return (rand() << 16) + rand(); }
const int MO = 1000000007;
const int N = 1000000 + 5;
int a[N];
int b[N];
int n, k;
long long L;
vector<int> u;
vector<int> e;
int dp[N][2];
int d[N];
int add(int a, int b) {
  a += b;
  if (a >= MO) a -= MO;
  return a;
}
int mul(int a, int b) { return ((long long)a * b) % MO; }
int E, mn;
int ans;
int m;
long long q;
void F(int j) {
  if (q >= j) ans = add(ans, mul((q - j + 1) % MO, dp[m - 1][mn]));
  if (q >= j - 1 && E) {
    if (j == 1) {
      for (int i = 0; i < (int)(m); ++i) ans = add(ans, e[i]);
    } else {
      for (int i = 0; i < (int)(m); ++i)
        ans = add(ans, mul(e[i], dp[i][mn ^ 1]));
    }
  }
}
void solve() {
  cin >> n >> L >> k;
  for (int i = 0; i < (int)(n); ++i) scanf("%d", &a[i]);
  q = L / n;
  vector<pair<int, int> > v(n);
  for (int i = 0; i < (int)(n); ++i) v[i] = make_pair(a[i], i);
  sort((v).begin(), (v).end());
  for (int i = 0; i < (int)(n); ++i) b[i] = v[i].first;
  E = L % n;
  u.push_back(1);
  e.push_back(0);
  if (v[0].second < E) {
    ++e.back();
  }
  for (int i = 1; i < n; ++i) {
    if (b[i] != b[i - 1]) {
      u.push_back(0);
      e.push_back(0);
    }
    ++u.back();
    if (v[i].second < E) ++e.back();
  }
  m = ((int)((e).size()));
  mn = 0;
  for (int i = 0; i < (int)(m); ++i) d[i] = u[i];
  dp[0][mn] = d[0];
  for (int i = 1; i <= (int)(m - 1); ++i) dp[i][mn] = add(d[i], dp[i - 1][mn]);
  ans = 0;
  F(1);
  for (int j = 2; j <= k; ++j) {
    int nm = mn ^ 1;
    for (int i = 0; i < (int)(m); ++i) d[i] = mul(u[i], dp[i][mn]);
    mn = nm;
    dp[0][mn] = d[0];
    for (int i = 1; i <= (int)(m - 1); ++i)
      dp[i][mn] = add(d[i], dp[i - 1][mn]);
    F(j);
  }
  printf("%d\n", ans);
}
void testgen() { FILE* f = fopen("input.txt", "w"); }
int main() {
  cout << fixed;
  cout.precision(16);
  cerr << fixed;
  cerr.precision(3);
  solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, "/STACK:66777216")
using namespace std;
int RAND() { return (rand() << 16) + rand(); }
const int MO = 1000000007;
const int N = 1000000 + 5;
int a[N];
int b[N];
int n, k;
long long L;
vector<int> u;
vector<int> e;
int dp[N][2];
int d[N];
int add(int a, int b) {
  a += b;
  if (a >= MO) a -= MO;
  return a;
}
int mul(int a, int b) { return ((long long)a * b) % MO; }
int E, mn;
int ans;
int m;
long long q;
void F(int j) {
  if (q >= j) ans = add(ans, mul((q - j + 1) % MO, dp[m - 1][mn]));
  if (q >= j - 1 && E) {
    if (j == 1) {
      for (int i = 0; i < (int)(m); ++i) ans = add(ans, e[i]);
    } else {
      for (int i = 0; i < (int)(m); ++i)
        ans = add(ans, mul(e[i], dp[i][mn ^ 1]));
    }
  }
}
void solve() {
  cin >> n >> L >> k;
  for (int i = 0; i < (int)(n); ++i) scanf("%d", &a[i]);
  q = L / n;
  vector<pair<int, int> > v(n);
  for (int i = 0; i < (int)(n); ++i) v[i] = make_pair(a[i], i);
  sort((v).begin(), (v).end());
  for (int i = 0; i < (int)(n); ++i) b[i] = v[i].first;
  E = L % n;
  u.push_back(1);
  e.push_back(0);
  if (v[0].second < E) {
    ++e.back();
  }
  for (int i = 1; i < n; ++i) {
    if (b[i] != b[i - 1]) {
      u.push_back(0);
      e.push_back(0);
    }
    ++u.back();
    if (v[i].second < E) ++e.back();
  }
  m = ((int)((e).size()));
  mn = 0;
  for (int i = 0; i < (int)(m); ++i) d[i] = u[i];
  dp[0][mn] = d[0];
  for (int i = 1; i <= (int)(m - 1); ++i) dp[i][mn] = add(d[i], dp[i - 1][mn]);
  ans = 0;
  F(1);
  for (int j = 2; j <= k; ++j) {
    int nm = mn ^ 1;
    for (int i = 0; i < (int)(m); ++i) d[i] = mul(u[i], dp[i][mn]);
    mn = nm;
    dp[0][mn] = d[0];
    for (int i = 1; i <= (int)(m - 1); ++i)
      dp[i][mn] = add(d[i], dp[i - 1][mn]);
    F(j);
  }
  printf("%d\n", ans);
}
void testgen() { FILE* f = fopen("input.txt", "w"); }
int main() {
  cout << fixed;
  cout.precision(16);
  cerr << fixed;
  cerr.precision(3);
  solve();
  return 0;
}

```