#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 5;
const int nax = 1e6 + 5;
const int mod = 1e9 + 7;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, lim;
long long total;
int t[nax];
void read() {
  scanf("%d%lld%d", &n, &total, &lim);
  set<int> s;
  for (int i = (0); i <= ((n)-1); ++i) {
    scanf("%d", &t[i]);
    s.insert(t[i]);
  }
  map<int, int> m;
  int i = 0;
  for (int a : s) m[a] = i++;
  for (int i = (0); i <= ((n)-1); ++i) t[i] = m[t[i]];
}
vector<vector<int> > dp;
int pre[nax];
int s[nax], s2[nax];
int main() {
  read();
  vector<int> single = vector<int>(n, 0);
  dp = vector<vector<int> >(lim + 1, single);
  dp[0][0] = 1;
  for (int row = (1); row <= ((lim)); ++row) {
    for (int i = (0); i <= ((n)-1); ++i) {
      pre[i] = dp[row - 1][i];
      if (i) add(pre[i], pre[i - 1]);
    }
    for (int i = (0); i <= ((n)-1); ++i) {
      add(dp[row][t[i]], pre[t[i]]);
      add(s[row], pre[t[i]]);
    }
  }
  for (int row = (1); row <= ((lim)); ++row) {
    for (int i = (0); i <= ((n)-1); ++i) {
      pre[i] = dp[row - 1][i];
      if (i) add(pre[i], pre[i - 1]);
    }
    int pom = total % n;
    for (int i = (0); i <= ((pom)-1); ++i) add(s2[row], pre[t[i]]);
  }
  int res = 0;
  long long full = total / n;
  for (int row = (1); row <= ((lim)); ++row) {
    long long x = (row <= full + 1);
    add(res, x * s2[row] % mod);
    x = (full - row + 1) % mod;
    if (x < 0) x = 0;
    add(res, x * s[row] % mod);
  }
  printf("%d\n", res);
  return 0;
}

### 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 namespace std;
const int inf = 1e9 + 5;
const int nax = 1e6 + 5;
const int mod = 1e9 + 7;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, lim;
long long total;
int t[nax];
void read() {
  scanf("%d%lld%d", &n, &total, &lim);
  set<int> s;
  for (int i = (0); i <= ((n)-1); ++i) {
    scanf("%d", &t[i]);
    s.insert(t[i]);
  }
  map<int, int> m;
  int i = 0;
  for (int a : s) m[a] = i++;
  for (int i = (0); i <= ((n)-1); ++i) t[i] = m[t[i]];
}
vector<vector<int> > dp;
int pre[nax];
int s[nax], s2[nax];
int main() {
  read();
  vector<int> single = vector<int>(n, 0);
  dp = vector<vector<int> >(lim + 1, single);
  dp[0][0] = 1;
  for (int row = (1); row <= ((lim)); ++row) {
    for (int i = (0); i <= ((n)-1); ++i) {
      pre[i] = dp[row - 1][i];
      if (i) add(pre[i], pre[i - 1]);
    }
    for (int i = (0); i <= ((n)-1); ++i) {
      add(dp[row][t[i]], pre[t[i]]);
      add(s[row], pre[t[i]]);
    }
  }
  for (int row = (1); row <= ((lim)); ++row) {
    for (int i = (0); i <= ((n)-1); ++i) {
      pre[i] = dp[row - 1][i];
      if (i) add(pre[i], pre[i - 1]);
    }
    int pom = total % n;
    for (int i = (0); i <= ((pom)-1); ++i) add(s2[row], pre[t[i]]);
  }
  int res = 0;
  long long full = total / n;
  for (int row = (1); row <= ((lim)); ++row) {
    long long x = (row <= full + 1);
    add(res, x * s2[row] % mod);
    x = (full - row + 1) % mod;
    if (x < 0) x = 0;
    add(res, x * s[row] % mod);
  }
  printf("%d\n", res);
  return 0;
}

```