#include <bits/stdc++.h>
using namespace std;
const int DI[] = {-1, 0, 1, 0};
const int DJ[] = {0, 1, 0, -1};
const int MAXINT = 0x7fffffff;
const int BIGINT = 0x20000000;
const long long MAXLL = 0x7fffffffffffffffLL;
const long long BIGLL = 0x2000000000000000LL;
const long long MOD = 1000000007;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T powmod(T a, T b, T m) {
  T r = 1;
  a %= m;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
  }
  return r;
}
long long n, l, k;
long long last, cgroup, group;
vector<pair<int, int> > a;
void equalify(vector<int>& dp) {
  for (int i = 0; i + 1 < n; ++i) {
    if (a[i].first == a[i + 1].first) dp[i + 1] = dp[i];
  }
}
int main() {
  scanf("%lld %lld %lld", &n, &l, &k);
  last = l % n, cgroup = l / n, group = (l - 1) / n + 1;
  a.reserve(n);
  for (int i = 0; i < n; ++i) {
    int num;
    scanf("%d", &num);
    a.push_back(pair<int, int>(num, i));
  }
  sort(a.begin(), a.end());
  vector<int> dp(n, 1), dp_tail(n);
  for (int i = 0; i < n; ++i) dp_tail[i] = a[i].second < last;
  long long ret = 0;
  for (long long t = 1; t <= min(cgroup, k); ++t) {
    for (int i = n - 2; i >= 0; --i) dp[i] = (dp[i] + dp[i + 1]) % MOD;
    equalify(dp);
    ret = (ret + dp[0] * ((cgroup - t + 1) % MOD) % MOD) % MOD;
  }
  for (long long t = 1; t <= min(group, k); ++t) {
    for (int i = n - 2; i >= 0; --i)
      dp_tail[i] = (dp_tail[i] + dp_tail[i + 1]) % MOD;
    equalify(dp_tail);
    ret = (ret + dp_tail[0]) % MOD;
  }
  printf("%d\n", (int)ret);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 DI[] = {-1, 0, 1, 0};
const int DJ[] = {0, 1, 0, -1};
const int MAXINT = 0x7fffffff;
const int BIGINT = 0x20000000;
const long long MAXLL = 0x7fffffffffffffffLL;
const long long BIGLL = 0x2000000000000000LL;
const long long MOD = 1000000007;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T powmod(T a, T b, T m) {
  T r = 1;
  a %= m;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
  }
  return r;
}
long long n, l, k;
long long last, cgroup, group;
vector<pair<int, int> > a;
void equalify(vector<int>& dp) {
  for (int i = 0; i + 1 < n; ++i) {
    if (a[i].first == a[i + 1].first) dp[i + 1] = dp[i];
  }
}
int main() {
  scanf("%lld %lld %lld", &n, &l, &k);
  last = l % n, cgroup = l / n, group = (l - 1) / n + 1;
  a.reserve(n);
  for (int i = 0; i < n; ++i) {
    int num;
    scanf("%d", &num);
    a.push_back(pair<int, int>(num, i));
  }
  sort(a.begin(), a.end());
  vector<int> dp(n, 1), dp_tail(n);
  for (int i = 0; i < n; ++i) dp_tail[i] = a[i].second < last;
  long long ret = 0;
  for (long long t = 1; t <= min(cgroup, k); ++t) {
    for (int i = n - 2; i >= 0; --i) dp[i] = (dp[i] + dp[i + 1]) % MOD;
    equalify(dp);
    ret = (ret + dp[0] * ((cgroup - t + 1) % MOD) % MOD) % MOD;
  }
  for (long long t = 1; t <= min(group, k); ++t) {
    for (int i = n - 2; i >= 0; --i)
      dp_tail[i] = (dp_tail[i] + dp_tail[i + 1]) % MOD;
    equalify(dp_tail);
    ret = (ret + dp_tail[0]) % MOD;
  }
  printf("%d\n", (int)ret);
}

```