#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const long long INF = 2147483647;
const long long LINF = 9223372036854775807;
const long long MAXN = 1e6 + 100, MOD = 1e9 + 7;
inline void EnableFileIO(const string& fileName) {
  if (fileName.empty()) return;
  freopen((fileName + ".in").c_str(), "r", stdin);
  freopen((fileName + ".out").c_str(), "w", stdout);
  return;
}
long long n, k;
long long l;
pair<long long, long long> a[MAXN];
long long rest[MAXN], pos[MAXN], dp[MAXN], sum[MAXN];
long long ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  EnableFileIO("");
  cin >> n >> l >> k;
  for (long long i = 0; i < n; i++) cin >> a[i].first, a[i].second = i;
  sort(a, a + n);
  for (long long i = 0, j = 0; i < n; i++) {
    while (j < n && a[j].first <= a[i].first) j++;
    pos[i] = j;
    for (long long j = i + n; j < n * k; j += n) a[j] = a[i];
  }
  for (long long i = 0; i < n; i++) dp[i] = 1;
  ans = l % MOD;
  for (long long i = 1; i < k; i++) {
    for (long long j = 1; j <= n; j++) sum[j] = (sum[j - 1] + dp[j - 1]) % MOD;
    for (long long j = 0; j < n; j++) {
      dp[j] = sum[pos[j]] % MOD;
      if (l / n + (a[j].second < l % n) < i) continue;
      ans += (l / n + (a[j].second < l % n) - i) % MOD * dp[j] % MOD;
      ans %= MOD;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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>
#pragma GCC optimize("O3")
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const long long INF = 2147483647;
const long long LINF = 9223372036854775807;
const long long MAXN = 1e6 + 100, MOD = 1e9 + 7;
inline void EnableFileIO(const string& fileName) {
  if (fileName.empty()) return;
  freopen((fileName + ".in").c_str(), "r", stdin);
  freopen((fileName + ".out").c_str(), "w", stdout);
  return;
}
long long n, k;
long long l;
pair<long long, long long> a[MAXN];
long long rest[MAXN], pos[MAXN], dp[MAXN], sum[MAXN];
long long ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  EnableFileIO("");
  cin >> n >> l >> k;
  for (long long i = 0; i < n; i++) cin >> a[i].first, a[i].second = i;
  sort(a, a + n);
  for (long long i = 0, j = 0; i < n; i++) {
    while (j < n && a[j].first <= a[i].first) j++;
    pos[i] = j;
    for (long long j = i + n; j < n * k; j += n) a[j] = a[i];
  }
  for (long long i = 0; i < n; i++) dp[i] = 1;
  ans = l % MOD;
  for (long long i = 1; i < k; i++) {
    for (long long j = 1; j <= n; j++) sum[j] = (sum[j - 1] + dp[j - 1]) % MOD;
    for (long long j = 0; j < n; j++) {
      dp[j] = sum[pos[j]] % MOD;
      if (l / n + (a[j].second < l % n) < i) continue;
      ans += (l / n + (a[j].second < l % n) - i) % MOD * dp[j] % MOD;
      ans %= MOD;
    }
  }
  cout << ans << endl;
  return 0;
}

```